Electric Potential

Electric potential energy, voltage, and calculating potential from fields

Electric Potential

Electric Potential Energy

Work done by electric force:

W=∫ABF⃗⋅dl⃗=q∫ABE⃗⋅dl⃗W = \int_A^B \vec{F} \cdot d\vec{l} = q\int_A^B \vec{E} \cdot d\vec{l}

Potential energy change: ΔU=−W=−q∫ABE⃗⋅dl⃗\Delta U = -W = -q\int_A^B \vec{E} \cdot d\vec{l}

Electric Potential (Voltage)

V=UqV = \frac{U}{q}

Potential difference: ΔV=VB−VA=−∫ABE⃗⋅dl⃗\Delta V = V_B - V_A = -\int_A^B \vec{E} \cdot d\vec{l}

(Path-independent for electrostatic fields)

Units: 1 volt = 1 joule/coulomb

Electric Field from Potential

E⃗=−∇V=−(∂V∂xi^+∂V∂yj^+∂V∂zk^)\vec{E} = -\nabla V = -\left(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k}\right)

In one dimension: Ex=−dVdxE_x = -\frac{dV}{dx}

Field points from high to low potential (down the potential gradient).

Potential of Point Charge

Choose V=0V = 0 at r=∞r = \infty:

V(r)=−∫∞rE⃗⋅dl⃗=−∫∞rkqr2 drV(r) = -\int_\infty^r \vec{E} \cdot d\vec{l} = -\int_\infty^r k\frac{q}{r^2} \, dr

V(r)=kqrV(r) = k\frac{q}{r}

For multiple charges: V=∑ikqiriV = \sum_i k\frac{q_i}{r_i}

(Scalar sum, easier than vector sum for E⃗\vec{E}!)

Potential of Continuous Distributions

V=k∫dqrV = k\int \frac{dq}{r}

Infinite Line Charge

V=−λ2πϵ0ln⁡rr0V = -\frac{\lambda}{2\pi\epsilon_0}\ln\frac{r}{r_0}

(Must choose reference point r0r_0 since V→∞V \to \infty at infinity)

Ring of Charge

On axis at distance xx:

V=kQx2+R2V = \frac{kQ}{\sqrt{x^2 + R^2}}

At center: V=kQ/RV = kQ/R

Disk of Charge

On axis:

V=σ2ϵ0(x2+R2−x)V = \frac{\sigma}{2\epsilon_0}\left(\sqrt{x^2 + R^2} - x\right)

Spherical Shell

Total charge QQ, radius RR:

Outside (r>Rr > R): V=kQ/rV = kQ/r

On surface (r=Rr = R): V=kQ/RV = kQ/R

Inside (r<Rr < R): V=kQ/RV = kQ/R (constant!)

Solid Sphere

Uniform charge density, total QQ, radius RR:

Outside (r>Rr > R): V=kQ/rV = kQ/r

Inside (r<Rr < R): V=kQ2R3(3R2−r2)V = \frac{kQ}{2R^3}(3R^2 - r^2)

At center: V=3kQ/(2R)V = 3kQ/(2R)

Equipotential Surfaces

Surfaces where VV = constant.

  • Electric field perpendicular to equipotentials
  • No work to move charge along equipotential
  • Conductors are equipotentials

Electric Dipole Potential

Far field (r≫dr \gg d):

V=kpcos⁡θr2V = k\frac{p\cos\theta}{r^2}

where θ\theta is angle from dipole axis and p=qdp = qd.

Electric field: Er=−∂V∂r=2kpcos⁡θr3E_r = -\frac{\partial V}{\partial r} = \frac{2kp\cos\theta}{r^3}

Eθ=−1r∂V∂θ=kpsin⁡θr3E_\theta = -\frac{1}{r}\frac{\partial V}{\partial \theta} = \frac{kp\sin\theta}{r^3}

Energy of Charge Distributions

Work to assemble charges:

U=12∑iqiViU = \frac{1}{2}\sum_{i} q_i V_i

where ViV_i is potential at location of qiq_i due to all other charges.

For continuous distribution: U=ϵ02∫E2 dVU = \frac{\epsilon_0}{2}\int E^2 \, dV

(Energy stored in electric field)

📚 Practice Problems

1Problem 1medium

❓ Question:

A uniformly charged ring of radius R = 0.15 m carries total charge Q = 8.0 × 10⁻⁹ C. (a) Find the electric potential at a point on the axis at distance x = 0.20 m from the center. (b) Find the electric field at this point from E = -dV/dx. (c) Find the work done to bring a charge q = 2.0 × 10⁻⁹ C from infinity to this point.

💡 Show Solution

Given:

  • R = 0.15 m
  • Q = 8.0 × 10⁻⁹ C = 8.0 nC
  • x = 0.20 m
  • q = 2.0 × 10⁻⁹ C = 2.0 nC
  • k = 8.99 × 10⁹ N·m²/C²

(a) Potential on axis:

All points on ring are at same distance from point P: r=x2+R2=(0.20)2+(0.15)2r = \sqrt{x^2 + R^2} = \sqrt{(0.20)^2 + (0.15)^2}

r=0.04+0.0225=0.0625=0.25 mr = \sqrt{0.04 + 0.0225} = \sqrt{0.0625} = 0.25 \text{ m}

Potential (scalar sum): V=kQr=kQx2+R2V = \frac{kQ}{r} = \frac{kQ}{\sqrt{x^2 + R^2}}

V=(8.99×109)(8.0×10−9)0.25V = \frac{(8.99 \times 10^9)(8.0 \times 10^{-9})}{0.25}

V=71.920.25=287.68 VV = \frac{71.92}{0.25} = 287.68 \text{ V}

(b) Electric field from potential:

Ex=−dVdx=−ddx[kQx2+R2]E_x = -\frac{dV}{dx} = -\frac{d}{dx}\left[\frac{kQ}{\sqrt{x^2 + R^2}}\right]

=−kQ⋅(−12)(x2+R2)−3/2(2x)= -kQ \cdot \left(-\frac{1}{2}\right)(x^2 + R^2)^{-3/2}(2x)

Ex=kQx(x2+R2)3/2E_x = \frac{kQx}{(x^2 + R^2)^{3/2}}

Ex=(8.99×109)(8.0×10−9)(0.20)(0.25)3E_x = \frac{(8.99 \times 10^9)(8.0 \times 10^{-9})(0.20)}{(0.25)^3}

Ex=14.3840.015625=920.6 N/CE_x = \frac{14.384}{0.015625} = 920.6 \text{ N/C}

(c) Work done:

Work to bring charge from infinity: W=qΔV=q(Vf−Vi)=q(V−0)W = qΔV = q(V_f - V_i) = q(V - 0)

W=(2.0×10−9)(287.68)W = (2.0 \times 10^{-9})(287.68)

W=5.75×10−7 J=575 nJW = 5.75 \times 10^{-7} \text{ J} = 575 \text{ nJ}

Note: Work is positive because we're bringing like charges together (both positive).

Answers: (a) V = 288 V (b) E = 921 N/C (along axis, away from ring) (c) W = 575 nJ

2Problem 2medium

❓ Question:

A uniformly charged ring of radius R = 0.15 m carries total charge Q = 8.0 × 10⁻⁹ C. (a) Find the electric potential at a point on the axis at distance x = 0.20 m from the center. (b) Find the electric field at this point from E = -dV/dx. (c) Find the work done to bring a charge q = 2.0 × 10⁻⁹ C from infinity to this point.

💡 Show Solution

Given:

  • R = 0.15 m
  • Q = 8.0 × 10⁻⁹ C = 8.0 nC
  • x = 0.20 m
  • q = 2.0 × 10⁻⁹ C = 2.0 nC
  • k = 8.99 × 10⁹ N·m²/C²

(a) Potential on axis:

All points on ring are at same distance from point P: r=x2+R2=(0.20)2+(0.15)2r = \sqrt{x^2 + R^2} = \sqrt{(0.20)^2 + (0.15)^2}

r=0.04+0.0225=0.0625=0.25 mr = \sqrt{0.04 + 0.0225} = \sqrt{0.0625} = 0.25 \text{ m}

Potential (scalar sum): V=kQr=kQx2+R2V = \frac{kQ}{r} = \frac{kQ}{\sqrt{x^2 + R^2}}

V=(8.99×109)(8.0×10−9)0.25V = \frac{(8.99 \times 10^9)(8.0 \times 10^{-9})}{0.25}

V=71.920.25=287.68 VV = \frac{71.92}{0.25} = 287.68 \text{ V}

(b) Electric field from potential:

Ex=−dVdx=−ddx[kQx2+R2]E_x = -\frac{dV}{dx} = -\frac{d}{dx}\left[\frac{kQ}{\sqrt{x^2 + R^2}}\right]

=−kQ⋅(−12)(x2+R2)−3/2(2x)= -kQ \cdot \left(-\frac{1}{2}\right)(x^2 + R^2)^{-3/2}(2x)

Ex=kQx(x2+R2)3/2E_x = \frac{kQx}{(x^2 + R^2)^{3/2}}

Ex=(8.99×109)(8.0×10−9)(0.20)(0.25)3E_x = \frac{(8.99 \times 10^9)(8.0 \times 10^{-9})(0.20)}{(0.25)^3}

Ex=14.3840.015625=920.6 N/CE_x = \frac{14.384}{0.015625} = 920.6 \text{ N/C}

(c) Work done:

Work to bring charge from infinity: W=qΔV=q(Vf−Vi)=q(V−0)W = qΔV = q(V_f - V_i) = q(V - 0)

W=(2.0×10−9)(287.68)W = (2.0 \times 10^{-9})(287.68)

W=5.75×10−7 J=575 nJW = 5.75 \times 10^{-7} \text{ J} = 575 \text{ nJ}

Note: Work is positive because we're bringing like charges together (both positive).

Answers: (a) V = 288 V (b) E = 921 N/C (along axis, away from ring) (c) W = 575 nJ

3Problem 3medium

❓ Question:

A uniformly charged ring of radius R = 0.15 m carries total charge Q = 8.0 × 10⁻⁹ C. (a) Find the electric potential at a point on the axis at distance x = 0.20 m from the center. (b) Find the electric field at this point from E = -dV/dx. (c) Find the work done to bring a charge q = 2.0 × 10⁻⁹ C from infinity to this point.

💡 Show Solution

Given:

  • R = 0.15 m
  • Q = 8.0 × 10⁻⁹ C = 8.0 nC
  • x = 0.20 m
  • q = 2.0 × 10⁻⁹ C = 2.0 nC
  • k = 8.99 × 10⁹ N·m²/C²

(a) Potential on axis:

All points on ring are at same distance from point P: r=x2+R2=(0.20)2+(0.15)2r = \sqrt{x^2 + R^2} = \sqrt{(0.20)^2 + (0.15)^2}

r=0.04+0.0225=0.0625=0.25 mr = \sqrt{0.04 + 0.0225} = \sqrt{0.0625} = 0.25 \text{ m}

Potential (scalar sum): V=kQr=kQx2+R2V = \frac{kQ}{r} = \frac{kQ}{\sqrt{x^2 + R^2}}

V=(8.99×109)(8.0×10−9)0.25V = \frac{(8.99 \times 10^9)(8.0 \times 10^{-9})}{0.25}

V=71.920.25=287.68 VV = \frac{71.92}{0.25} = 287.68 \text{ V}

(b) Electric field from potential:

Ex=−dVdx=−ddx[kQx2+R2]E_x = -\frac{dV}{dx} = -\frac{d}{dx}\left[\frac{kQ}{\sqrt{x^2 + R^2}}\right]

=−kQ⋅(−12)(x2+R2)−3/2(2x)= -kQ \cdot \left(-\frac{1}{2}\right)(x^2 + R^2)^{-3/2}(2x)

Ex=kQx(x2+R2)3/2E_x = \frac{kQx}{(x^2 + R^2)^{3/2}}

Ex=(8.99×109)(8.0×10−9)(0.20)(0.25)3E_x = \frac{(8.99 \times 10^9)(8.0 \times 10^{-9})(0.20)}{(0.25)^3}

Ex=14.3840.015625=920.6 N/CE_x = \frac{14.384}{0.015625} = 920.6 \text{ N/C}

(c) Work done:

Work to bring charge from infinity: W=qΔV=q(Vf−Vi)=q(V−0)W = qΔV = q(V_f - V_i) = q(V - 0)

W=(2.0×10−9)(287.68)W = (2.0 \times 10^{-9})(287.68)

W=5.75×10−7 J=575 nJW = 5.75 \times 10^{-7} \text{ J} = 575 \text{ nJ}

Note: Work is positive because we're bringing like charges together (both positive).

Answers: (a) V = 288 V (b) E = 921 N/C (along axis, away from ring) (c) W = 575 nJ

4Problem 4medium

❓ Question:

A uniformly charged ring of radius R = 0.15 m carries total charge Q = 8.0 × 10⁻⁹ C. (a) Find the electric potential at a point on the axis at distance x = 0.20 m from the center. (b) Find the electric field at this point from E = -dV/dx. (c) Find the work done to bring a charge q = 2.0 × 10⁻⁹ C from infinity to this point.

💡 Show Solution

Given:

  • R = 0.15 m
  • Q = 8.0 × 10⁻⁹ C = 8.0 nC
  • x = 0.20 m
  • q = 2.0 × 10⁻⁹ C = 2.0 nC
  • k = 8.99 × 10⁹ N·m²/C²

(a) Potential on axis:

All points on ring are at same distance from point P: r=x2+R2=(0.20)2+(0.15)2r = \sqrt{x^2 + R^2} = \sqrt{(0.20)^2 + (0.15)^2}

r=0.04+0.0225=0.0625=0.25 mr = \sqrt{0.04 + 0.0225} = \sqrt{0.0625} = 0.25 \text{ m}

Potential (scalar sum): V=kQr=kQx2+R2V = \frac{kQ}{r} = \frac{kQ}{\sqrt{x^2 + R^2}}

V=(8.99×109)(8.0×10−9)0.25V = \frac{(8.99 \times 10^9)(8.0 \times 10^{-9})}{0.25}

V=71.920.25=287.68 VV = \frac{71.92}{0.25} = 287.68 \text{ V}

(b) Electric field from potential:

Ex=−dVdx=−ddx[kQx2+R2]E_x = -\frac{dV}{dx} = -\frac{d}{dx}\left[\frac{kQ}{\sqrt{x^2 + R^2}}\right]

=−kQ⋅(−12)(x2+R2)−3/2(2x)= -kQ \cdot \left(-\frac{1}{2}\right)(x^2 + R^2)^{-3/2}(2x)

Ex=kQx(x2+R2)3/2E_x = \frac{kQx}{(x^2 + R^2)^{3/2}}

Ex=(8.99×109)(8.0×10−9)(0.20)(0.25)3E_x = \frac{(8.99 \times 10^9)(8.0 \times 10^{-9})(0.20)}{(0.25)^3}

Ex=14.3840.015625=920.6 N/CE_x = \frac{14.384}{0.015625} = 920.6 \text{ N/C}

(c) Work done:

Work to bring charge from infinity: W=qΔV=q(Vf−Vi)=q(V−0)W = qΔV = q(V_f - V_i) = q(V - 0)

W=(2.0×10−9)(287.68)W = (2.0 \times 10^{-9})(287.68)

W=5.75×10−7 J=575 nJW = 5.75 \times 10^{-7} \text{ J} = 575 \text{ nJ}

Note: Work is positive because we're bringing like charges together (both positive).

Answers: (a) V = 288 V (b) E = 921 N/C (along axis, away from ring) (c) W = 575 nJ

5Problem 5medium

❓ Question:

A uniformly charged ring of radius R = 0.15 m carries total charge Q = 8.0 × 10⁻⁹ C. (a) Find the electric potential at a point on the axis at distance x = 0.20 m from the center. (b) Find the electric field at this point from E = -dV/dx. (c) Find the work done to bring a charge q = 2.0 × 10⁻⁹ C from infinity to this point.

💡 Show Solution

Given:

  • R = 0.15 m
  • Q = 8.0 × 10⁻⁹ C = 8.0 nC
  • x = 0.20 m
  • q = 2.0 × 10⁻⁹ C = 2.0 nC
  • k = 8.99 × 10⁹ N·m²/C²

(a) Potential on axis:

All points on ring are at same distance from point P: r=x2+R2=(0.20)2+(0.15)2r = \sqrt{x^2 + R^2} = \sqrt{(0.20)^2 + (0.15)^2}

r=0.04+0.0225=0.0625=0.25 mr = \sqrt{0.04 + 0.0225} = \sqrt{0.0625} = 0.25 \text{ m}

Potential (scalar sum): V=kQr=kQx2+R2V = \frac{kQ}{r} = \frac{kQ}{\sqrt{x^2 + R^2}}

V=(8.99×109)(8.0×10−9)0.25V = \frac{(8.99 \times 10^9)(8.0 \times 10^{-9})}{0.25}

V=71.920.25=287.68 VV = \frac{71.92}{0.25} = 287.68 \text{ V}

(b) Electric field from potential:

Ex=−dVdx=−ddx[kQx2+R2]E_x = -\frac{dV}{dx} = -\frac{d}{dx}\left[\frac{kQ}{\sqrt{x^2 + R^2}}\right]

=−kQ⋅(−12)(x2+R2)−3/2(2x)= -kQ \cdot \left(-\frac{1}{2}\right)(x^2 + R^2)^{-3/2}(2x)

Ex=kQx(x2+R2)3/2E_x = \frac{kQx}{(x^2 + R^2)^{3/2}}

Ex=(8.99×109)(8.0×10−9)(0.20)(0.25)3E_x = \frac{(8.99 \times 10^9)(8.0 \times 10^{-9})(0.20)}{(0.25)^3}

Ex=14.3840.015625=920.6 N/CE_x = \frac{14.384}{0.015625} = 920.6 \text{ N/C}

(c) Work done:

Work to bring charge from infinity: W=qΔV=q(Vf−Vi)=q(V−0)W = qΔV = q(V_f - V_i) = q(V - 0)

W=(2.0×10−9)(287.68)W = (2.0 \times 10^{-9})(287.68)

W=5.75×10−7 J=575 nJW = 5.75 \times 10^{-7} \text{ J} = 575 \text{ nJ}

Note: Work is positive because we're bringing like charges together (both positive).

Answers: (a) V = 288 V (b) E = 921 N/C (along axis, away from ring) (c) W = 575 nJ

6Problem 6medium

❓ Question:

A uniformly charged ring of radius R = 0.15 m carries total charge Q = 8.0 × 10⁻⁹ C. (a) Find the electric potential at a point on the axis at distance x = 0.20 m from the center. (b) Find the electric field at this point from E = -dV/dx. (c) Find the work done to bring a charge q = 2.0 × 10⁻⁹ C from infinity to this point.

💡 Show Solution

Given:

  • R = 0.15 m
  • Q = 8.0 × 10⁻⁹ C = 8.0 nC
  • x = 0.20 m
  • q = 2.0 × 10⁻⁹ C = 2.0 nC
  • k = 8.99 × 10⁹ N·m²/C²

(a) Potential on axis:

All points on ring are at same distance from point P: r=x2+R2=(0.20)2+(0.15)2r = \sqrt{x^2 + R^2} = \sqrt{(0.20)^2 + (0.15)^2}

r=0.04+0.0225=0.0625=0.25 mr = \sqrt{0.04 + 0.0225} = \sqrt{0.0625} = 0.25 \text{ m}

Potential (scalar sum): V=kQr=kQx2+R2V = \frac{kQ}{r} = \frac{kQ}{\sqrt{x^2 + R^2}}

V=(8.99×109)(8.0×10−9)0.25V = \frac{(8.99 \times 10^9)(8.0 \times 10^{-9})}{0.25}

V=71.920.25=287.68 VV = \frac{71.92}{0.25} = 287.68 \text{ V}

(b) Electric field from potential:

Ex=−dVdx=−ddx[kQx2+R2]E_x = -\frac{dV}{dx} = -\frac{d}{dx}\left[\frac{kQ}{\sqrt{x^2 + R^2}}\right]

=−kQ⋅(−12)(x2+R2)−3/2(2x)= -kQ \cdot \left(-\frac{1}{2}\right)(x^2 + R^2)^{-3/2}(2x)

Ex=kQx(x2+R2)3/2E_x = \frac{kQx}{(x^2 + R^2)^{3/2}}

Ex=(8.99×109)(8.0×10−9)(0.20)(0.25)3E_x = \frac{(8.99 \times 10^9)(8.0 \times 10^{-9})(0.20)}{(0.25)^3}

Ex=14.3840.015625=920.6 N/CE_x = \frac{14.384}{0.015625} = 920.6 \text{ N/C}

(c) Work done:

Work to bring charge from infinity: W=qΔV=q(Vf−Vi)=q(V−0)W = qΔV = q(V_f - V_i) = q(V - 0)

W=(2.0×10−9)(287.68)W = (2.0 \times 10^{-9})(287.68)

W=5.75×10−7 J=575 nJW = 5.75 \times 10^{-7} \text{ J} = 575 \text{ nJ}

Note: Work is positive because we're bringing like charges together (both positive).

Answers: (a) V = 288 V (b) E = 921 N/C (along axis, away from ring) (c) W = 575 nJ

7Problem 7medium

❓ Question:

A uniformly charged solid sphere of radius R = 0.10 m has total charge Q = 6.0 × 10⁻⁸ C. Find the electric potential: (a) at the center (r = 0), (b) at the surface (r = R), and (c) at a point outside at r = 0.20 m.

💡 Show Solution

Given:

  • R = 0.10 m
  • Q = 6.0 × 10⁻⁸ C = 60 nC
  • k = 8.99 × 10⁹ N·m²/C²

Formula for uniformly charged solid sphere:

Outside (r ≥ R): V(r)=kQrV(r) = \frac{kQ}{r}

Inside (r < R): V(r)=kQ2R3(3R2−r2)V(r) = \frac{kQ}{2R^3}(3R^2 - r^2)

(a) Potential at center (r = 0):

V(0)=kQ2R3(3R2−0)=3kQ2RV(0) = \frac{kQ}{2R^3}(3R^2 - 0) = \frac{3kQ}{2R}

V(0)=3(8.99×109)(6.0×10−8)2(0.10)V(0) = \frac{3(8.99 \times 10^9)(6.0 \times 10^{-8})}{2(0.10)}

V(0)=1618.20.20=8091 VV(0) = \frac{1618.2}{0.20} = 8091 \text{ V}

(b) Potential at surface (r = R):

Using inside formula: V(R)=kQ2R3(3R2−R2)=kQ2R3(2R2)=kQRV(R) = \frac{kQ}{2R^3}(3R^2 - R^2) = \frac{kQ}{2R^3}(2R^2) = \frac{kQ}{R}

Or using outside formula: V(R)=kQRV(R) = \frac{kQ}{R}

V(R)=(8.99×109)(6.0×10−8)0.10V(R) = \frac{(8.99 \times 10^9)(6.0 \times 10^{-8})}{0.10}

V(R)=5394 VV(R) = 5394 \text{ V}

(c) Potential at r = 0.20 m > R:

V(0.20)=kQr=(8.99×109)(6.0×10−8)0.20V(0.20) = \frac{kQ}{r} = \frac{(8.99 \times 10^9)(6.0 \times 10^{-8})}{0.20}

V(0.20)=2697 VV(0.20) = 2697 \text{ V}

Verification:

  • V(0) = (3/2)V(R) = (3/2)(5394) = 8091 V ✓
  • V(0.20) = (1/2)V(R) since r = 2R ✓

Potential ratio: V(0)V(R)=3/2⋅kQ/RkQ/R=32=1.5\frac{V(0)}{V(R)} = \frac{3/2 \cdot kQ/R}{kQ/R} = \frac{3}{2} = 1.5

Answers: (a) V(0) = 8.09 kV (center) (b) V(R) = 5.39 kV (surface) (c) V(0.20 m) = 2.70 kV (outside)

8Problem 8medium

❓ Question:

A uniformly charged solid sphere of radius R = 0.10 m has total charge Q = 6.0 × 10⁻⁸ C. Find the electric potential: (a) at the center (r = 0), (b) at the surface (r = R), and (c) at a point outside at r = 0.20 m.

💡 Show Solution

Given:

  • R = 0.10 m
  • Q = 6.0 × 10⁻⁸ C = 60 nC
  • k = 8.99 × 10⁹ N·m²/C²

Formula for uniformly charged solid sphere:

Outside (r ≥ R): V(r)=kQrV(r) = \frac{kQ}{r}

Inside (r < R): V(r)=kQ2R3(3R2−r2)V(r) = \frac{kQ}{2R^3}(3R^2 - r^2)

(a) Potential at center (r = 0):

V(0)=kQ2R3(3R2−0)=3kQ2RV(0) = \frac{kQ}{2R^3}(3R^2 - 0) = \frac{3kQ}{2R}

V(0)=3(8.99×109)(6.0×10−8)2(0.10)V(0) = \frac{3(8.99 \times 10^9)(6.0 \times 10^{-8})}{2(0.10)}

V(0)=1618.20.20=8091 VV(0) = \frac{1618.2}{0.20} = 8091 \text{ V}

(b) Potential at surface (r = R):

Using inside formula: V(R)=kQ2R3(3R2−R2)=kQ2R3(2R2)=kQRV(R) = \frac{kQ}{2R^3}(3R^2 - R^2) = \frac{kQ}{2R^3}(2R^2) = \frac{kQ}{R}

Or using outside formula: V(R)=kQRV(R) = \frac{kQ}{R}

V(R)=(8.99×109)(6.0×10−8)0.10V(R) = \frac{(8.99 \times 10^9)(6.0 \times 10^{-8})}{0.10}

V(R)=5394 VV(R) = 5394 \text{ V}

(c) Potential at r = 0.20 m > R:

V(0.20)=kQr=(8.99×109)(6.0×10−8)0.20V(0.20) = \frac{kQ}{r} = \frac{(8.99 \times 10^9)(6.0 \times 10^{-8})}{0.20}

V(0.20)=2697 VV(0.20) = 2697 \text{ V}

Verification:

  • V(0) = (3/2)V(R) = (3/2)(5394) = 8091 V ✓
  • V(0.20) = (1/2)V(R) since r = 2R ✓

Potential ratio: V(0)V(R)=3/2⋅kQ/RkQ/R=32=1.5\frac{V(0)}{V(R)} = \frac{3/2 \cdot kQ/R}{kQ/R} = \frac{3}{2} = 1.5

Answers: (a) V(0) = 8.09 kV (center) (b) V(R) = 5.39 kV (surface) (c) V(0.20 m) = 2.70 kV (outside)

9Problem 9medium

❓ Question:

A uniformly charged solid sphere of radius R = 0.10 m has total charge Q = 6.0 × 10⁻⁸ C. Find the electric potential: (a) at the center (r = 0), (b) at the surface (r = R), and (c) at a point outside at r = 0.20 m.

💡 Show Solution

Given:

  • R = 0.10 m
  • Q = 6.0 × 10⁻⁸ C = 60 nC
  • k = 8.99 × 10⁹ N·m²/C²

Formula for uniformly charged solid sphere:

Outside (r ≥ R): V(r)=kQrV(r) = \frac{kQ}{r}

Inside (r < R): V(r)=kQ2R3(3R2−r2)V(r) = \frac{kQ}{2R^3}(3R^2 - r^2)

(a) Potential at center (r = 0):

V(0)=kQ2R3(3R2−0)=3kQ2RV(0) = \frac{kQ}{2R^3}(3R^2 - 0) = \frac{3kQ}{2R}

V(0)=3(8.99×109)(6.0×10−8)2(0.10)V(0) = \frac{3(8.99 \times 10^9)(6.0 \times 10^{-8})}{2(0.10)}

V(0)=1618.20.20=8091 VV(0) = \frac{1618.2}{0.20} = 8091 \text{ V}

(b) Potential at surface (r = R):

Using inside formula: V(R)=kQ2R3(3R2−R2)=kQ2R3(2R2)=kQRV(R) = \frac{kQ}{2R^3}(3R^2 - R^2) = \frac{kQ}{2R^3}(2R^2) = \frac{kQ}{R}

Or using outside formula: V(R)=kQRV(R) = \frac{kQ}{R}

V(R)=(8.99×109)(6.0×10−8)0.10V(R) = \frac{(8.99 \times 10^9)(6.0 \times 10^{-8})}{0.10}

V(R)=5394 VV(R) = 5394 \text{ V}

(c) Potential at r = 0.20 m > R:

V(0.20)=kQr=(8.99×109)(6.0×10−8)0.20V(0.20) = \frac{kQ}{r} = \frac{(8.99 \times 10^9)(6.0 \times 10^{-8})}{0.20}

V(0.20)=2697 VV(0.20) = 2697 \text{ V}

Verification:

  • V(0) = (3/2)V(R) = (3/2)(5394) = 8091 V ✓
  • V(0.20) = (1/2)V(R) since r = 2R ✓

Potential ratio: V(0)V(R)=3/2⋅kQ/RkQ/R=32=1.5\frac{V(0)}{V(R)} = \frac{3/2 \cdot kQ/R}{kQ/R} = \frac{3}{2} = 1.5

Answers: (a) V(0) = 8.09 kV (center) (b) V(R) = 5.39 kV (surface) (c) V(0.20 m) = 2.70 kV (outside)

10Problem 10medium

❓ Question:

A uniformly charged solid sphere of radius R = 0.10 m has total charge Q = 6.0 × 10⁻⁸ C. Find the electric potential: (a) at the center (r = 0), (b) at the surface (r = R), and (c) at a point outside at r = 0.20 m.

💡 Show Solution

Given:

  • R = 0.10 m
  • Q = 6.0 × 10⁻⁸ C = 60 nC
  • k = 8.99 × 10⁹ N·m²/C²

Formula for uniformly charged solid sphere:

Outside (r ≥ R): V(r)=kQrV(r) = \frac{kQ}{r}

Inside (r < R): V(r)=kQ2R3(3R2−r2)V(r) = \frac{kQ}{2R^3}(3R^2 - r^2)

(a) Potential at center (r = 0):

V(0)=kQ2R3(3R2−0)=3kQ2RV(0) = \frac{kQ}{2R^3}(3R^2 - 0) = \frac{3kQ}{2R}

V(0)=3(8.99×109)(6.0×10−8)2(0.10)V(0) = \frac{3(8.99 \times 10^9)(6.0 \times 10^{-8})}{2(0.10)}

V(0)=1618.20.20=8091 VV(0) = \frac{1618.2}{0.20} = 8091 \text{ V}

(b) Potential at surface (r = R):

Using inside formula: V(R)=kQ2R3(3R2−R2)=kQ2R3(2R2)=kQRV(R) = \frac{kQ}{2R^3}(3R^2 - R^2) = \frac{kQ}{2R^3}(2R^2) = \frac{kQ}{R}

Or using outside formula: V(R)=kQRV(R) = \frac{kQ}{R}

V(R)=(8.99×109)(6.0×10−8)0.10V(R) = \frac{(8.99 \times 10^9)(6.0 \times 10^{-8})}{0.10}

V(R)=5394 VV(R) = 5394 \text{ V}

(c) Potential at r = 0.20 m > R:

V(0.20)=kQr=(8.99×109)(6.0×10−8)0.20V(0.20) = \frac{kQ}{r} = \frac{(8.99 \times 10^9)(6.0 \times 10^{-8})}{0.20}

V(0.20)=2697 VV(0.20) = 2697 \text{ V}

Verification:

  • V(0) = (3/2)V(R) = (3/2)(5394) = 8091 V ✓
  • V(0.20) = (1/2)V(R) since r = 2R ✓

Potential ratio: V(0)V(R)=3/2⋅kQ/RkQ/R=32=1.5\frac{V(0)}{V(R)} = \frac{3/2 \cdot kQ/R}{kQ/R} = \frac{3}{2} = 1.5

Answers: (a) V(0) = 8.09 kV (center) (b) V(R) = 5.39 kV (surface) (c) V(0.20 m) = 2.70 kV (outside)

11Problem 11medium

❓ Question:

A uniformly charged solid sphere of radius R = 0.10 m has total charge Q = 6.0 × 10⁻⁸ C. Find the electric potential: (a) at the center (r = 0), (b) at the surface (r = R), and (c) at a point outside at r = 0.20 m.

💡 Show Solution

Given:

  • R = 0.10 m
  • Q = 6.0 × 10⁻⁸ C = 60 nC
  • k = 8.99 × 10⁹ N·m²/C²

Formula for uniformly charged solid sphere:

Outside (r ≥ R): V(r)=kQrV(r) = \frac{kQ}{r}

Inside (r < R): V(r)=kQ2R3(3R2−r2)V(r) = \frac{kQ}{2R^3}(3R^2 - r^2)

(a) Potential at center (r = 0):

V(0)=kQ2R3(3R2−0)=3kQ2RV(0) = \frac{kQ}{2R^3}(3R^2 - 0) = \frac{3kQ}{2R}

V(0)=3(8.99×109)(6.0×10−8)2(0.10)V(0) = \frac{3(8.99 \times 10^9)(6.0 \times 10^{-8})}{2(0.10)}

V(0)=1618.20.20=8091 VV(0) = \frac{1618.2}{0.20} = 8091 \text{ V}

(b) Potential at surface (r = R):

Using inside formula: V(R)=kQ2R3(3R2−R2)=kQ2R3(2R2)=kQRV(R) = \frac{kQ}{2R^3}(3R^2 - R^2) = \frac{kQ}{2R^3}(2R^2) = \frac{kQ}{R}

Or using outside formula: V(R)=kQRV(R) = \frac{kQ}{R}

V(R)=(8.99×109)(6.0×10−8)0.10V(R) = \frac{(8.99 \times 10^9)(6.0 \times 10^{-8})}{0.10}

V(R)=5394 VV(R) = 5394 \text{ V}

(c) Potential at r = 0.20 m > R:

V(0.20)=kQr=(8.99×109)(6.0×10−8)0.20V(0.20) = \frac{kQ}{r} = \frac{(8.99 \times 10^9)(6.0 \times 10^{-8})}{0.20}

V(0.20)=2697 VV(0.20) = 2697 \text{ V}

Verification:

  • V(0) = (3/2)V(R) = (3/2)(5394) = 8091 V ✓
  • V(0.20) = (1/2)V(R) since r = 2R ✓

Potential ratio: V(0)V(R)=3/2⋅kQ/RkQ/R=32=1.5\frac{V(0)}{V(R)} = \frac{3/2 \cdot kQ/R}{kQ/R} = \frac{3}{2} = 1.5

Answers: (a) V(0) = 8.09 kV (center) (b) V(R) = 5.39 kV (surface) (c) V(0.20 m) = 2.70 kV (outside)

12Problem 12medium

❓ Question:

A uniformly charged solid sphere of radius R = 0.10 m has total charge Q = 6.0 × 10⁻⁸ C. Find the electric potential: (a) at the center (r = 0), (b) at the surface (r = R), and (c) at a point outside at r = 0.20 m.

💡 Show Solution

Given:

  • R = 0.10 m
  • Q = 6.0 × 10⁻⁸ C = 60 nC
  • k = 8.99 × 10⁹ N·m²/C²

Formula for uniformly charged solid sphere:

Outside (r ≥ R): V(r)=kQrV(r) = \frac{kQ}{r}

Inside (r < R): V(r)=kQ2R3(3R2−r2)V(r) = \frac{kQ}{2R^3}(3R^2 - r^2)

(a) Potential at center (r = 0):

V(0)=kQ2R3(3R2−0)=3kQ2RV(0) = \frac{kQ}{2R^3}(3R^2 - 0) = \frac{3kQ}{2R}

V(0)=3(8.99×109)(6.0×10−8)2(0.10)V(0) = \frac{3(8.99 \times 10^9)(6.0 \times 10^{-8})}{2(0.10)}

V(0)=1618.20.20=8091 VV(0) = \frac{1618.2}{0.20} = 8091 \text{ V}

(b) Potential at surface (r = R):

Using inside formula: V(R)=kQ2R3(3R2−R2)=kQ2R3(2R2)=kQRV(R) = \frac{kQ}{2R^3}(3R^2 - R^2) = \frac{kQ}{2R^3}(2R^2) = \frac{kQ}{R}

Or using outside formula: V(R)=kQRV(R) = \frac{kQ}{R}

V(R)=(8.99×109)(6.0×10−8)0.10V(R) = \frac{(8.99 \times 10^9)(6.0 \times 10^{-8})}{0.10}

V(R)=5394 VV(R) = 5394 \text{ V}

(c) Potential at r = 0.20 m > R:

V(0.20)=kQr=(8.99×109)(6.0×10−8)0.20V(0.20) = \frac{kQ}{r} = \frac{(8.99 \times 10^9)(6.0 \times 10^{-8})}{0.20}

V(0.20)=2697 VV(0.20) = 2697 \text{ V}

Verification:

  • V(0) = (3/2)V(R) = (3/2)(5394) = 8091 V ✓
  • V(0.20) = (1/2)V(R) since r = 2R ✓

Potential ratio: V(0)V(R)=3/2⋅kQ/RkQ/R=32=1.5\frac{V(0)}{V(R)} = \frac{3/2 \cdot kQ/R}{kQ/R} = \frac{3}{2} = 1.5

Answers: (a) V(0) = 8.09 kV (center) (b) V(R) = 5.39 kV (surface) (c) V(0.20 m) = 2.70 kV (outside)

13Problem 13hard

❓ Question:

Three point charges are located at the vertices of an equilateral triangle with side length a = 0.30 m. The charges are q₁ = +4.0 nC, q₂ = +4.0 nC, and q₃ = -4.0 nC. Find: (a) the electric potential at the center of the triangle, (b) the electric field at the center, and (c) the work required to bring a charge q = +2.0 nC from infinity to the center.

💡 Show Solution

Given:

  • a = 0.30 m
  • q₁ = q₂ = +4.0 × 10⁻⁹ C
  • q₃ = -4.0 × 10⁻⁹ C
  • q = +2.0 × 10⁻⁹ C
  • k = 8.99 × 10⁹ N·m²/C²

Distance from vertices to center:

For equilateral triangle, center to vertex: r=a3=0.303=0.173 mr = \frac{a}{\sqrt{3}} = \frac{0.30}{\sqrt{3}} = 0.173 \text{ m}

(a) Potential at center:

Potential is scalar sum: V=k∑qiri=kr(q1+q2+q3)V = k\sum \frac{q_i}{r_i} = \frac{k}{r}(q_1 + q_2 + q_3)

V=8.99×1090.173(4.0×10−9+4.0×10−9−4.0×10−9)V = \frac{8.99 \times 10^9}{0.173}(4.0 \times 10^{-9} + 4.0 \times 10^{-9} - 4.0 \times 10^{-9})

V=8.99×1090.173(4.0×10−9)V = \frac{8.99 \times 10^9}{0.173}(4.0 \times 10^{-9})

V=(5.20×1010)(4.0×10−9)=208 VV = (5.20 \times 10^{10})(4.0 \times 10^{-9}) = 208 \text{ V}

(b) Electric field at center:

Set up coordinates: q₁ at top, q₂ at bottom-left, q₃ at bottom-right

Field from each charge: Ei=kqir2=(8.99×109)(4.0×10−9)(0.173)2E_i = \frac{kq_i}{r^2} = \frac{(8.99 \times 10^9)(4.0 \times 10^{-9})}{(0.173)^2}

Ei=35.960.0299=1203 N/CE_i = \frac{35.96}{0.0299} = 1203 \text{ N/C}

Vector addition:

By symmetry, q₁ and q₂ (both positive, same magnitude):

  • Their fields point away from charges
  • The horizontal components cancel
  • Vertical components add

The field from q₃ (negative) points toward it (upward by our setup).

Due to symmetry of two positive charges and one negative of equal magnitude:

Net field components:

  • E₁ points down from top: E₁ = 1203 N/C (down)
  • E₂ points from bottom-left: at 60° from horizontal
  • E₃ points toward bottom-right: at 60° from horizontal (opposite side)

E₂ and E₃ have components that partially cancel...

Actually, with two +4 nC and one -4 nC arranged symmetrically, the net field has magnitude:

Enet=3k∣q∣r2=3(1203)=3609 N/CE_{net} = \frac{3k|q|}{r^2} = 3(1203) = 3609 \text{ N/C}

Direction: toward the negative charge.

(c) Work to bring charge from infinity:

W=qV=(2.0×10−9)(208)=4.16×10−7 JW = qV = (2.0 \times 10^{-9})(208) = 4.16 \times 10^{-7} \text{ J}

W=416 nJW = 416 \text{ nJ}

Answers: (a) V = 208 V (b) E ≈ 3.6 kN/C (toward q₃) (c) W = 416 nJ

14Problem 14hard

❓ Question:

Three point charges are located at the vertices of an equilateral triangle with side length a = 0.30 m. The charges are q₁ = +4.0 nC, q₂ = +4.0 nC, and q₃ = -4.0 nC. Find: (a) the electric potential at the center of the triangle, (b) the electric field at the center, and (c) the work required to bring a charge q = +2.0 nC from infinity to the center.

💡 Show Solution

Given:

  • a = 0.30 m
  • q₁ = q₂ = +4.0 × 10⁻⁹ C
  • q₃ = -4.0 × 10⁻⁹ C
  • q = +2.0 × 10⁻⁹ C
  • k = 8.99 × 10⁹ N·m²/C²

Distance from vertices to center:

For equilateral triangle, center to vertex: r=a3=0.303=0.173 mr = \frac{a}{\sqrt{3}} = \frac{0.30}{\sqrt{3}} = 0.173 \text{ m}

(a) Potential at center:

Potential is scalar sum: V=k∑qiri=kr(q1+q2+q3)V = k\sum \frac{q_i}{r_i} = \frac{k}{r}(q_1 + q_2 + q_3)

V=8.99×1090.173(4.0×10−9+4.0×10−9−4.0×10−9)V = \frac{8.99 \times 10^9}{0.173}(4.0 \times 10^{-9} + 4.0 \times 10^{-9} - 4.0 \times 10^{-9})

V=8.99×1090.173(4.0×10−9)V = \frac{8.99 \times 10^9}{0.173}(4.0 \times 10^{-9})

V=(5.20×1010)(4.0×10−9)=208 VV = (5.20 \times 10^{10})(4.0 \times 10^{-9}) = 208 \text{ V}

(b) Electric field at center:

Set up coordinates: q₁ at top, q₂ at bottom-left, q₃ at bottom-right

Field from each charge: Ei=kqir2=(8.99×109)(4.0×10−9)(0.173)2E_i = \frac{kq_i}{r^2} = \frac{(8.99 \times 10^9)(4.0 \times 10^{-9})}{(0.173)^2}

Ei=35.960.0299=1203 N/CE_i = \frac{35.96}{0.0299} = 1203 \text{ N/C}

Vector addition:

By symmetry, q₁ and q₂ (both positive, same magnitude):

  • Their fields point away from charges
  • The horizontal components cancel
  • Vertical components add

The field from q₃ (negative) points toward it (upward by our setup).

Due to symmetry of two positive charges and one negative of equal magnitude:

Net field components:

  • E₁ points down from top: E₁ = 1203 N/C (down)
  • E₂ points from bottom-left: at 60° from horizontal
  • E₃ points toward bottom-right: at 60° from horizontal (opposite side)

E₂ and E₃ have components that partially cancel...

Actually, with two +4 nC and one -4 nC arranged symmetrically, the net field has magnitude:

Enet=3k∣q∣r2=3(1203)=3609 N/CE_{net} = \frac{3k|q|}{r^2} = 3(1203) = 3609 \text{ N/C}

Direction: toward the negative charge.

(c) Work to bring charge from infinity:

W=qV=(2.0×10−9)(208)=4.16×10−7 JW = qV = (2.0 \times 10^{-9})(208) = 4.16 \times 10^{-7} \text{ J}

W=416 nJW = 416 \text{ nJ}

Answers: (a) V = 208 V (b) E ≈ 3.6 kN/C (toward q₃) (c) W = 416 nJ

15Problem 15hard

❓ Question:

Three point charges are located at the vertices of an equilateral triangle with side length a = 0.30 m. The charges are q₁ = +4.0 nC, q₂ = +4.0 nC, and q₃ = -4.0 nC. Find: (a) the electric potential at the center of the triangle, (b) the electric field at the center, and (c) the work required to bring a charge q = +2.0 nC from infinity to the center.

💡 Show Solution

Given:

  • a = 0.30 m
  • q₁ = q₂ = +4.0 × 10⁻⁹ C
  • q₃ = -4.0 × 10⁻⁹ C
  • q = +2.0 × 10⁻⁹ C
  • k = 8.99 × 10⁹ N·m²/C²

Distance from vertices to center:

For equilateral triangle, center to vertex: r=a3=0.303=0.173 mr = \frac{a}{\sqrt{3}} = \frac{0.30}{\sqrt{3}} = 0.173 \text{ m}

(a) Potential at center:

Potential is scalar sum: V=k∑qiri=kr(q1+q2+q3)V = k\sum \frac{q_i}{r_i} = \frac{k}{r}(q_1 + q_2 + q_3)

V=8.99×1090.173(4.0×10−9+4.0×10−9−4.0×10−9)V = \frac{8.99 \times 10^9}{0.173}(4.0 \times 10^{-9} + 4.0 \times 10^{-9} - 4.0 \times 10^{-9})

V=8.99×1090.173(4.0×10−9)V = \frac{8.99 \times 10^9}{0.173}(4.0 \times 10^{-9})

V=(5.20×1010)(4.0×10−9)=208 VV = (5.20 \times 10^{10})(4.0 \times 10^{-9}) = 208 \text{ V}

(b) Electric field at center:

Set up coordinates: q₁ at top, q₂ at bottom-left, q₃ at bottom-right

Field from each charge: Ei=kqir2=(8.99×109)(4.0×10−9)(0.173)2E_i = \frac{kq_i}{r^2} = \frac{(8.99 \times 10^9)(4.0 \times 10^{-9})}{(0.173)^2}

Ei=35.960.0299=1203 N/CE_i = \frac{35.96}{0.0299} = 1203 \text{ N/C}

Vector addition:

By symmetry, q₁ and q₂ (both positive, same magnitude):

  • Their fields point away from charges
  • The horizontal components cancel
  • Vertical components add

The field from q₃ (negative) points toward it (upward by our setup).

Due to symmetry of two positive charges and one negative of equal magnitude:

Net field components:

  • E₁ points down from top: E₁ = 1203 N/C (down)
  • E₂ points from bottom-left: at 60° from horizontal
  • E₃ points toward bottom-right: at 60° from horizontal (opposite side)

E₂ and E₃ have components that partially cancel...

Actually, with two +4 nC and one -4 nC arranged symmetrically, the net field has magnitude:

Enet=3k∣q∣r2=3(1203)=3609 N/CE_{net} = \frac{3k|q|}{r^2} = 3(1203) = 3609 \text{ N/C}

Direction: toward the negative charge.

(c) Work to bring charge from infinity:

W=qV=(2.0×10−9)(208)=4.16×10−7 JW = qV = (2.0 \times 10^{-9})(208) = 4.16 \times 10^{-7} \text{ J}

W=416 nJW = 416 \text{ nJ}

Answers: (a) V = 208 V (b) E ≈ 3.6 kN/C (toward q₃) (c) W = 416 nJ

16Problem 16hard

❓ Question:

Three point charges are located at the vertices of an equilateral triangle with side length a = 0.30 m. The charges are q₁ = +4.0 nC, q₂ = +4.0 nC, and q₃ = -4.0 nC. Find: (a) the electric potential at the center of the triangle, (b) the electric field at the center, and (c) the work required to bring a charge q = +2.0 nC from infinity to the center.

💡 Show Solution

Given:

  • a = 0.30 m
  • q₁ = q₂ = +4.0 × 10⁻⁹ C
  • q₃ = -4.0 × 10⁻⁹ C
  • q = +2.0 × 10⁻⁹ C
  • k = 8.99 × 10⁹ N·m²/C²

Distance from vertices to center:

For equilateral triangle, center to vertex: r=a3=0.303=0.173 mr = \frac{a}{\sqrt{3}} = \frac{0.30}{\sqrt{3}} = 0.173 \text{ m}

(a) Potential at center:

Potential is scalar sum: V=k∑qiri=kr(q1+q2+q3)V = k\sum \frac{q_i}{r_i} = \frac{k}{r}(q_1 + q_2 + q_3)

V=8.99×1090.173(4.0×10−9+4.0×10−9−4.0×10−9)V = \frac{8.99 \times 10^9}{0.173}(4.0 \times 10^{-9} + 4.0 \times 10^{-9} - 4.0 \times 10^{-9})

V=8.99×1090.173(4.0×10−9)V = \frac{8.99 \times 10^9}{0.173}(4.0 \times 10^{-9})

V=(5.20×1010)(4.0×10−9)=208 VV = (5.20 \times 10^{10})(4.0 \times 10^{-9}) = 208 \text{ V}

(b) Electric field at center:

Set up coordinates: q₁ at top, q₂ at bottom-left, q₃ at bottom-right

Field from each charge: Ei=kqir2=(8.99×109)(4.0×10−9)(0.173)2E_i = \frac{kq_i}{r^2} = \frac{(8.99 \times 10^9)(4.0 \times 10^{-9})}{(0.173)^2}

Ei=35.960.0299=1203 N/CE_i = \frac{35.96}{0.0299} = 1203 \text{ N/C}

Vector addition:

By symmetry, q₁ and q₂ (both positive, same magnitude):

  • Their fields point away from charges
  • The horizontal components cancel
  • Vertical components add

The field from q₃ (negative) points toward it (upward by our setup).

Due to symmetry of two positive charges and one negative of equal magnitude:

Net field components:

  • E₁ points down from top: E₁ = 1203 N/C (down)
  • E₂ points from bottom-left: at 60° from horizontal
  • E₃ points toward bottom-right: at 60° from horizontal (opposite side)

E₂ and E₃ have components that partially cancel...

Actually, with two +4 nC and one -4 nC arranged symmetrically, the net field has magnitude:

Enet=3k∣q∣r2=3(1203)=3609 N/CE_{net} = \frac{3k|q|}{r^2} = 3(1203) = 3609 \text{ N/C}

Direction: toward the negative charge.

(c) Work to bring charge from infinity:

W=qV=(2.0×10−9)(208)=4.16×10−7 JW = qV = (2.0 \times 10^{-9})(208) = 4.16 \times 10^{-7} \text{ J}

W=416 nJW = 416 \text{ nJ}

Answers: (a) V = 208 V (b) E ≈ 3.6 kN/C (toward q₃) (c) W = 416 nJ

17Problem 17hard

❓ Question:

Three point charges are located at the vertices of an equilateral triangle with side length a = 0.30 m. The charges are q₁ = +4.0 nC, q₂ = +4.0 nC, and q₃ = -4.0 nC. Find: (a) the electric potential at the center of the triangle, (b) the electric field at the center, and (c) the work required to bring a charge q = +2.0 nC from infinity to the center.

💡 Show Solution

Given:

  • a = 0.30 m
  • q₁ = q₂ = +4.0 × 10⁻⁹ C
  • q₃ = -4.0 × 10⁻⁹ C
  • q = +2.0 × 10⁻⁹ C
  • k = 8.99 × 10⁹ N·m²/C²

Distance from vertices to center:

For equilateral triangle, center to vertex: r=a3=0.303=0.173 mr = \frac{a}{\sqrt{3}} = \frac{0.30}{\sqrt{3}} = 0.173 \text{ m}

(a) Potential at center:

Potential is scalar sum: V=k∑qiri=kr(q1+q2+q3)V = k\sum \frac{q_i}{r_i} = \frac{k}{r}(q_1 + q_2 + q_3)

V=8.99×1090.173(4.0×10−9+4.0×10−9−4.0×10−9)V = \frac{8.99 \times 10^9}{0.173}(4.0 \times 10^{-9} + 4.0 \times 10^{-9} - 4.0 \times 10^{-9})

V=8.99×1090.173(4.0×10−9)V = \frac{8.99 \times 10^9}{0.173}(4.0 \times 10^{-9})

V=(5.20×1010)(4.0×10−9)=208 VV = (5.20 \times 10^{10})(4.0 \times 10^{-9}) = 208 \text{ V}

(b) Electric field at center:

Set up coordinates: q₁ at top, q₂ at bottom-left, q₃ at bottom-right

Field from each charge: Ei=kqir2=(8.99×109)(4.0×10−9)(0.173)2E_i = \frac{kq_i}{r^2} = \frac{(8.99 \times 10^9)(4.0 \times 10^{-9})}{(0.173)^2}

Ei=35.960.0299=1203 N/CE_i = \frac{35.96}{0.0299} = 1203 \text{ N/C}

Vector addition:

By symmetry, q₁ and q₂ (both positive, same magnitude):

  • Their fields point away from charges
  • The horizontal components cancel
  • Vertical components add

The field from q₃ (negative) points toward it (upward by our setup).

Due to symmetry of two positive charges and one negative of equal magnitude:

Net field components:

  • E₁ points down from top: E₁ = 1203 N/C (down)
  • E₂ points from bottom-left: at 60° from horizontal
  • E₃ points toward bottom-right: at 60° from horizontal (opposite side)

E₂ and E₃ have components that partially cancel...

Actually, with two +4 nC and one -4 nC arranged symmetrically, the net field has magnitude:

Enet=3k∣q∣r2=3(1203)=3609 N/CE_{net} = \frac{3k|q|}{r^2} = 3(1203) = 3609 \text{ N/C}

Direction: toward the negative charge.

(c) Work to bring charge from infinity:

W=qV=(2.0×10−9)(208)=4.16×10−7 JW = qV = (2.0 \times 10^{-9})(208) = 4.16 \times 10^{-7} \text{ J}

W=416 nJW = 416 \text{ nJ}

Answers: (a) V = 208 V (b) E ≈ 3.6 kN/C (toward q₃) (c) W = 416 nJ

18Problem 18hard

❓ Question:

Three point charges are located at the vertices of an equilateral triangle with side length a = 0.30 m. The charges are q₁ = +4.0 nC, q₂ = +4.0 nC, and q₃ = -4.0 nC. Find: (a) the electric potential at the center of the triangle, (b) the electric field at the center, and (c) the work required to bring a charge q = +2.0 nC from infinity to the center.

💡 Show Solution

Given:

  • a = 0.30 m
  • q₁ = q₂ = +4.0 × 10⁻⁹ C
  • q₃ = -4.0 × 10⁻⁹ C
  • q = +2.0 × 10⁻⁹ C
  • k = 8.99 × 10⁹ N·m²/C²

Distance from vertices to center:

For equilateral triangle, center to vertex: r=a3=0.303=0.173 mr = \frac{a}{\sqrt{3}} = \frac{0.30}{\sqrt{3}} = 0.173 \text{ m}

(a) Potential at center:

Potential is scalar sum: V=k∑qiri=kr(q1+q2+q3)V = k\sum \frac{q_i}{r_i} = \frac{k}{r}(q_1 + q_2 + q_3)

V=8.99×1090.173(4.0×10−9+4.0×10−9−4.0×10−9)V = \frac{8.99 \times 10^9}{0.173}(4.0 \times 10^{-9} + 4.0 \times 10^{-9} - 4.0 \times 10^{-9})

V=8.99×1090.173(4.0×10−9)V = \frac{8.99 \times 10^9}{0.173}(4.0 \times 10^{-9})

V=(5.20×1010)(4.0×10−9)=208 VV = (5.20 \times 10^{10})(4.0 \times 10^{-9}) = 208 \text{ V}

(b) Electric field at center:

Set up coordinates: q₁ at top, q₂ at bottom-left, q₃ at bottom-right

Field from each charge: Ei=kqir2=(8.99×109)(4.0×10−9)(0.173)2E_i = \frac{kq_i}{r^2} = \frac{(8.99 \times 10^9)(4.0 \times 10^{-9})}{(0.173)^2}

Ei=35.960.0299=1203 N/CE_i = \frac{35.96}{0.0299} = 1203 \text{ N/C}

Vector addition:

By symmetry, q₁ and q₂ (both positive, same magnitude):

  • Their fields point away from charges
  • The horizontal components cancel
  • Vertical components add

The field from q₃ (negative) points toward it (upward by our setup).

Due to symmetry of two positive charges and one negative of equal magnitude:

Net field components:

  • E₁ points down from top: E₁ = 1203 N/C (down)
  • E₂ points from bottom-left: at 60° from horizontal
  • E₃ points toward bottom-right: at 60° from horizontal (opposite side)

E₂ and E₃ have components that partially cancel...

Actually, with two +4 nC and one -4 nC arranged symmetrically, the net field has magnitude:

Enet=3k∣q∣r2=3(1203)=3609 N/CE_{net} = \frac{3k|q|}{r^2} = 3(1203) = 3609 \text{ N/C}

Direction: toward the negative charge.

(c) Work to bring charge from infinity:

W=qV=(2.0×10−9)(208)=4.16×10−7 JW = qV = (2.0 \times 10^{-9})(208) = 4.16 \times 10^{-7} \text{ J}

W=416 nJW = 416 \text{ nJ}

Answers: (a) V = 208 V (b) E ≈ 3.6 kN/C (toward q₃) (c) W = 416 nJ

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