Electric Field and Coulomb's Law

Coulomb's law, electric field from continuous charge distributions

Electric Field and Coulomb's Law

Coulomb's Law

Force between two point charges:

F⃗=kq1q2r2r^=14πϵ0q1q2r2r^\vec{F} = k\frac{q_1q_2}{r^2}\hat{r} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\hat{r}

where:

  • k=8.99×109k = 8.99 \times 10^9 N·m²/C²
  • ϵ0=8.85×10−12\epsilon_0 = 8.85 \times 10^{-12} C²/(N·m²) (permittivity of free space)

Vector form: F⃗12=kq1q2r122r^12\vec{F}_{12} = k\frac{q_1q_2}{r_{12}^2}\hat{r}_{12}

Electric Field

Electric field due to point charge:

E⃗=kqr2r^\vec{E} = k\frac{q}{r^2}\hat{r}

Definition: E⃗=F⃗q0\vec{E} = \frac{\vec{F}}{q_0}

where q0q_0 is test charge.

Superposition principle: E⃗total=∑iE⃗i\vec{E}_{total} = \sum_i \vec{E}_i

Continuous Charge Distributions

For continuous distribution with charge density ρ\rho:

E⃗=k∫dqr2r^\vec{E} = k\int \frac{dq}{r^2}\hat{r}

Linear Charge Density

Charge per unit length: λ=dq/dl\lambda = dq/dl

dE⃗=kλ dlr2r^d\vec{E} = k\frac{\lambda \, dl}{r^2}\hat{r}

Surface Charge Density

Charge per unit area: σ=dq/dA\sigma = dq/dA

dE⃗=kσ dAr2r^d\vec{E} = k\frac{\sigma \, dA}{r^2}\hat{r}

Volume Charge Density

Charge per unit volume: ρ=dq/dV\rho = dq/dV

dE⃗=kρ dVr2r^d\vec{E} = k\frac{\rho \, dV}{r^2}\hat{r}

Example: Infinite Line of Charge

Uniform line charge density λ\lambda, find field at distance rr:

By symmetry, field is radial. Consider element at distance zz:

dEx=kλ dz(r2+z2)rr2+z2dE_x = \frac{k\lambda \, dz}{(r^2 + z^2)}\frac{r}{\sqrt{r^2 + z^2}}

E=∫−∞∞kλr dz(r2+z2)3/2E = \int_{-\infty}^{\infty} \frac{k\lambda r \, dz}{(r^2 + z^2)^{3/2}}

Using z=rtan⁡θz = r\tan\theta:

E=2kλr=λ2πϵ0rE = \frac{2k\lambda}{r} = \frac{\lambda}{2\pi\epsilon_0 r}

Example: Ring of Charge

Ring of radius RR, total charge QQ, find field on axis at distance xx:

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

At center (x=0x = 0): E=0E = 0 (by symmetry)

Far from ring (x≫Rx \gg R): E≈kQ/x2E \approx kQ/x^2 (point charge)

Example: Disk of Charge

Uniform surface charge density σ\sigma, radius RR, field on axis:

Ex=σ2ϵ0(1−xx2+R2)E_x = \frac{\sigma}{2\epsilon_0}\left(1 - \frac{x}{\sqrt{x^2 + R^2}}\right)

At surface (x=0x = 0): E=σ/(2ϵ0)E = \sigma/(2\epsilon_0)

Far from disk (x≫Rx \gg R): E≈kπR2σ/x2=kQ/x2E \approx k\pi R^2\sigma/x^2 = kQ/x^2

Infinite sheet (R→∞R \to \infty): E=σ/(2ϵ0)E = \sigma/(2\epsilon_0) (uniform!)

Example: Spherical Shell

Uniform surface charge density, total charge QQ, radius RR:

Outside (r>Rr > R): E=kQ/r2E = kQ/r^2 (like point charge)

Inside (r<Rr < R): E=0E = 0

(From Gauss's law, derived below)

Dipole

Two charges +q+q and −q-q separated by distance dd:

Dipole moment: p⃗=qd⃗\vec{p} = q\vec{d}

(points from −q-q to +q+q)

Field on axis (far field, r≫dr \gg d): Eaxis=2kpr3E_{axis} = \frac{2kp}{r^3}

Field on perpendicular bisector: Eperp=kpr3E_{perp} = \frac{kp}{r^3}

📚 Practice Problems

1Problem 1hard

❓ Question:

A thin rod of length L = 0.5 m has a non-uniform linear charge density λ(x) = λ₀x, where λ₀ = 4.0 × 10⁻⁶ C/m² and x is measured from one end. Find the electric field at a point P located a distance d = 0.2 m from the end where x = 0, along the axis of the rod.

💡 Show Solution

Given:

  • L = 0.5 m
  • λ(x) = λ₀x where λ₀ = 4.0 × 10⁻⁶ C/m²
  • d = 0.2 m
  • k = 8.99 × 10⁹ N·m²/C²

Setup:

For a small element dx at position x from the origin, the charge is: dq=λ(x)dx=λ0x dxdq = λ(x)dx = λ_0 x \, dx

The electric field contribution at P (located at distance d from x = 0): dE=kdqr2=kλ0x dx(d+x)2dE = k\frac{dq}{r^2} = k\frac{λ_0 x \, dx}{(d + x)^2}

Integration:

E=∫0Lkλ0x(d+x)2 dx=kλ0∫0Lx(d+x)2 dxE = \int_0^L k\frac{λ_0 x}{(d + x)^2} \, dx = kλ_0 \int_0^L \frac{x}{(d + x)^2} \, dx

Using substitution u = d + x, du = dx, when x = 0, u = d; when x = L, u = d + L: E=kλ0∫dd+Lu−du2 du=kλ0∫dd+L(1u−du2)duE = kλ_0 \int_d^{d+L} \frac{u - d}{u^2} \, du = kλ_0 \int_d^{d+L} \left(\frac{1}{u} - \frac{d}{u^2}\right) du

E=kλ0[ln⁡u+du]dd+LE = kλ_0 \left[\ln u + \frac{d}{u}\right]_d^{d+L}

E=kλ0[ln⁡(d+L)−ln⁡d+dd+L−1]E = kλ_0 \left[\ln(d+L) - \ln d + \frac{d}{d+L} - 1\right]

E=kλ0[ln⁡(d+Ld)+dd+L−1]E = kλ_0 \left[\ln\left(\frac{d+L}{d}\right) + \frac{d}{d+L} - 1\right]

Numerical calculation: E=(8.99×109)(4.0×10−6)[ln⁡(0.70.2)+0.20.7−1]E = (8.99 \times 10^9)(4.0 \times 10^{-6})\left[\ln\left(\frac{0.7}{0.2}\right) + \frac{0.2}{0.7} - 1\right]

E=35,960[ln⁡(3.5)+0.286−1]E = 35,960 \left[\ln(3.5) + 0.286 - 1\right]

E=35,960[1.253+0.286−1]=35,960(0.539)E = 35,960 \left[1.253 + 0.286 - 1\right] = 35,960(0.539)

E=1.94×104 N/CE = 1.94 \times 10^4 \text{ N/C}

Answer: The electric field at point P is 1.94 × 10⁴ N/C pointing away from the rod.

2Problem 2hard

❓ Question:

A thin rod of length L = 0.5 m has a non-uniform linear charge density λ(x) = λ₀x, where λ₀ = 4.0 × 10⁻⁶ C/m² and x is measured from one end. Find the electric field at a point P located a distance d = 0.2 m from the end where x = 0, along the axis of the rod.

💡 Show Solution

Given:

  • L = 0.5 m
  • λ(x) = λ₀x where λ₀ = 4.0 × 10⁻⁶ C/m²
  • d = 0.2 m
  • k = 8.99 × 10⁹ N·m²/C²

Setup:

For a small element dx at position x from the origin, the charge is: dq=λ(x)dx=λ0x dxdq = λ(x)dx = λ_0 x \, dx

The electric field contribution at P (located at distance d from x = 0): dE=kdqr2=kλ0x dx(d+x)2dE = k\frac{dq}{r^2} = k\frac{λ_0 x \, dx}{(d + x)^2}

Integration:

E=∫0Lkλ0x(d+x)2 dx=kλ0∫0Lx(d+x)2 dxE = \int_0^L k\frac{λ_0 x}{(d + x)^2} \, dx = kλ_0 \int_0^L \frac{x}{(d + x)^2} \, dx

Using substitution u = d + x, du = dx, when x = 0, u = d; when x = L, u = d + L: E=kλ0∫dd+Lu−du2 du=kλ0∫dd+L(1u−du2)duE = kλ_0 \int_d^{d+L} \frac{u - d}{u^2} \, du = kλ_0 \int_d^{d+L} \left(\frac{1}{u} - \frac{d}{u^2}\right) du

E=kλ0[ln⁡u+du]dd+LE = kλ_0 \left[\ln u + \frac{d}{u}\right]_d^{d+L}

E=kλ0[ln⁡(d+L)−ln⁡d+dd+L−1]E = kλ_0 \left[\ln(d+L) - \ln d + \frac{d}{d+L} - 1\right]

E=kλ0[ln⁡(d+Ld)+dd+L−1]E = kλ_0 \left[\ln\left(\frac{d+L}{d}\right) + \frac{d}{d+L} - 1\right]

Numerical calculation: E=(8.99×109)(4.0×10−6)[ln⁡(0.70.2)+0.20.7−1]E = (8.99 \times 10^9)(4.0 \times 10^{-6})\left[\ln\left(\frac{0.7}{0.2}\right) + \frac{0.2}{0.7} - 1\right]

E=35,960[ln⁡(3.5)+0.286−1]E = 35,960 \left[\ln(3.5) + 0.286 - 1\right]

E=35,960[1.253+0.286−1]=35,960(0.539)E = 35,960 \left[1.253 + 0.286 - 1\right] = 35,960(0.539)

E=1.94×104 N/CE = 1.94 \times 10^4 \text{ N/C}

Answer: The electric field at point P is 1.94 × 10⁴ N/C pointing away from the rod.

3Problem 3hard

❓ Question:

A thin rod of length L = 0.5 m has a non-uniform linear charge density λ(x) = λ₀x, where λ₀ = 4.0 × 10⁻⁶ C/m² and x is measured from one end. Find the electric field at a point P located a distance d = 0.2 m from the end where x = 0, along the axis of the rod.

💡 Show Solution

Given:

  • L = 0.5 m
  • λ(x) = λ₀x where λ₀ = 4.0 × 10⁻⁶ C/m²
  • d = 0.2 m
  • k = 8.99 × 10⁹ N·m²/C²

Setup:

For a small element dx at position x from the origin, the charge is: dq=λ(x)dx=λ0x dxdq = λ(x)dx = λ_0 x \, dx

The electric field contribution at P (located at distance d from x = 0): dE=kdqr2=kλ0x dx(d+x)2dE = k\frac{dq}{r^2} = k\frac{λ_0 x \, dx}{(d + x)^2}

Integration:

E=∫0Lkλ0x(d+x)2 dx=kλ0∫0Lx(d+x)2 dxE = \int_0^L k\frac{λ_0 x}{(d + x)^2} \, dx = kλ_0 \int_0^L \frac{x}{(d + x)^2} \, dx

Using substitution u = d + x, du = dx, when x = 0, u = d; when x = L, u = d + L: E=kλ0∫dd+Lu−du2 du=kλ0∫dd+L(1u−du2)duE = kλ_0 \int_d^{d+L} \frac{u - d}{u^2} \, du = kλ_0 \int_d^{d+L} \left(\frac{1}{u} - \frac{d}{u^2}\right) du

E=kλ0[ln⁡u+du]dd+LE = kλ_0 \left[\ln u + \frac{d}{u}\right]_d^{d+L}

E=kλ0[ln⁡(d+L)−ln⁡d+dd+L−1]E = kλ_0 \left[\ln(d+L) - \ln d + \frac{d}{d+L} - 1\right]

E=kλ0[ln⁡(d+Ld)+dd+L−1]E = kλ_0 \left[\ln\left(\frac{d+L}{d}\right) + \frac{d}{d+L} - 1\right]

Numerical calculation: E=(8.99×109)(4.0×10−6)[ln⁡(0.70.2)+0.20.7−1]E = (8.99 \times 10^9)(4.0 \times 10^{-6})\left[\ln\left(\frac{0.7}{0.2}\right) + \frac{0.2}{0.7} - 1\right]

E=35,960[ln⁡(3.5)+0.286−1]E = 35,960 \left[\ln(3.5) + 0.286 - 1\right]

E=35,960[1.253+0.286−1]=35,960(0.539)E = 35,960 \left[1.253 + 0.286 - 1\right] = 35,960(0.539)

E=1.94×104 N/CE = 1.94 \times 10^4 \text{ N/C}

Answer: The electric field at point P is 1.94 × 10⁴ N/C pointing away from the rod.

4Problem 4hard

❓ Question:

A thin rod of length L = 0.5 m has a non-uniform linear charge density λ(x) = λ₀x, where λ₀ = 4.0 × 10⁻⁶ C/m² and x is measured from one end. Find the electric field at a point P located a distance d = 0.2 m from the end where x = 0, along the axis of the rod.

💡 Show Solution

Given:

  • L = 0.5 m
  • λ(x) = λ₀x where λ₀ = 4.0 × 10⁻⁶ C/m²
  • d = 0.2 m
  • k = 8.99 × 10⁹ N·m²/C²

Setup:

For a small element dx at position x from the origin, the charge is: dq=λ(x)dx=λ0x dxdq = λ(x)dx = λ_0 x \, dx

The electric field contribution at P (located at distance d from x = 0): dE=kdqr2=kλ0x dx(d+x)2dE = k\frac{dq}{r^2} = k\frac{λ_0 x \, dx}{(d + x)^2}

Integration:

E=∫0Lkλ0x(d+x)2 dx=kλ0∫0Lx(d+x)2 dxE = \int_0^L k\frac{λ_0 x}{(d + x)^2} \, dx = kλ_0 \int_0^L \frac{x}{(d + x)^2} \, dx

Using substitution u = d + x, du = dx, when x = 0, u = d; when x = L, u = d + L: E=kλ0∫dd+Lu−du2 du=kλ0∫dd+L(1u−du2)duE = kλ_0 \int_d^{d+L} \frac{u - d}{u^2} \, du = kλ_0 \int_d^{d+L} \left(\frac{1}{u} - \frac{d}{u^2}\right) du

E=kλ0[ln⁡u+du]dd+LE = kλ_0 \left[\ln u + \frac{d}{u}\right]_d^{d+L}

E=kλ0[ln⁡(d+L)−ln⁡d+dd+L−1]E = kλ_0 \left[\ln(d+L) - \ln d + \frac{d}{d+L} - 1\right]

E=kλ0[ln⁡(d+Ld)+dd+L−1]E = kλ_0 \left[\ln\left(\frac{d+L}{d}\right) + \frac{d}{d+L} - 1\right]

Numerical calculation: E=(8.99×109)(4.0×10−6)[ln⁡(0.70.2)+0.20.7−1]E = (8.99 \times 10^9)(4.0 \times 10^{-6})\left[\ln\left(\frac{0.7}{0.2}\right) + \frac{0.2}{0.7} - 1\right]

E=35,960[ln⁡(3.5)+0.286−1]E = 35,960 \left[\ln(3.5) + 0.286 - 1\right]

E=35,960[1.253+0.286−1]=35,960(0.539)E = 35,960 \left[1.253 + 0.286 - 1\right] = 35,960(0.539)

E=1.94×104 N/CE = 1.94 \times 10^4 \text{ N/C}

Answer: The electric field at point P is 1.94 × 10⁴ N/C pointing away from the rod.

5Problem 5hard

❓ Question:

A thin rod of length L = 0.5 m has a non-uniform linear charge density λ(x) = λ₀x, where λ₀ = 4.0 × 10⁻⁶ C/m² and x is measured from one end. Find the electric field at a point P located a distance d = 0.2 m from the end where x = 0, along the axis of the rod.

💡 Show Solution

Given:

  • L = 0.5 m
  • λ(x) = λ₀x where λ₀ = 4.0 × 10⁻⁶ C/m²
  • d = 0.2 m
  • k = 8.99 × 10⁹ N·m²/C²

Setup:

For a small element dx at position x from the origin, the charge is: dq=λ(x)dx=λ0x dxdq = λ(x)dx = λ_0 x \, dx

The electric field contribution at P (located at distance d from x = 0): dE=kdqr2=kλ0x dx(d+x)2dE = k\frac{dq}{r^2} = k\frac{λ_0 x \, dx}{(d + x)^2}

Integration:

E=∫0Lkλ0x(d+x)2 dx=kλ0∫0Lx(d+x)2 dxE = \int_0^L k\frac{λ_0 x}{(d + x)^2} \, dx = kλ_0 \int_0^L \frac{x}{(d + x)^2} \, dx

Using substitution u = d + x, du = dx, when x = 0, u = d; when x = L, u = d + L: E=kλ0∫dd+Lu−du2 du=kλ0∫dd+L(1u−du2)duE = kλ_0 \int_d^{d+L} \frac{u - d}{u^2} \, du = kλ_0 \int_d^{d+L} \left(\frac{1}{u} - \frac{d}{u^2}\right) du

E=kλ0[ln⁡u+du]dd+LE = kλ_0 \left[\ln u + \frac{d}{u}\right]_d^{d+L}

E=kλ0[ln⁡(d+L)−ln⁡d+dd+L−1]E = kλ_0 \left[\ln(d+L) - \ln d + \frac{d}{d+L} - 1\right]

E=kλ0[ln⁡(d+Ld)+dd+L−1]E = kλ_0 \left[\ln\left(\frac{d+L}{d}\right) + \frac{d}{d+L} - 1\right]

Numerical calculation: E=(8.99×109)(4.0×10−6)[ln⁡(0.70.2)+0.20.7−1]E = (8.99 \times 10^9)(4.0 \times 10^{-6})\left[\ln\left(\frac{0.7}{0.2}\right) + \frac{0.2}{0.7} - 1\right]

E=35,960[ln⁡(3.5)+0.286−1]E = 35,960 \left[\ln(3.5) + 0.286 - 1\right]

E=35,960[1.253+0.286−1]=35,960(0.539)E = 35,960 \left[1.253 + 0.286 - 1\right] = 35,960(0.539)

E=1.94×104 N/CE = 1.94 \times 10^4 \text{ N/C}

Answer: The electric field at point P is 1.94 × 10⁴ N/C pointing away from the rod.

6Problem 6hard

❓ Question:

A thin rod of length L = 0.5 m has a non-uniform linear charge density λ(x) = λ₀x, where λ₀ = 4.0 × 10⁻⁶ C/m² and x is measured from one end. Find the electric field at a point P located a distance d = 0.2 m from the end where x = 0, along the axis of the rod.

💡 Show Solution

Given:

  • L = 0.5 m
  • λ(x) = λ₀x where λ₀ = 4.0 × 10⁻⁶ C/m²
  • d = 0.2 m
  • k = 8.99 × 10⁹ N·m²/C²

Setup:

For a small element dx at position x from the origin, the charge is: dq=λ(x)dx=λ0x dxdq = λ(x)dx = λ_0 x \, dx

The electric field contribution at P (located at distance d from x = 0): dE=kdqr2=kλ0x dx(d+x)2dE = k\frac{dq}{r^2} = k\frac{λ_0 x \, dx}{(d + x)^2}

Integration:

E=∫0Lkλ0x(d+x)2 dx=kλ0∫0Lx(d+x)2 dxE = \int_0^L k\frac{λ_0 x}{(d + x)^2} \, dx = kλ_0 \int_0^L \frac{x}{(d + x)^2} \, dx

Using substitution u = d + x, du = dx, when x = 0, u = d; when x = L, u = d + L: E=kλ0∫dd+Lu−du2 du=kλ0∫dd+L(1u−du2)duE = kλ_0 \int_d^{d+L} \frac{u - d}{u^2} \, du = kλ_0 \int_d^{d+L} \left(\frac{1}{u} - \frac{d}{u^2}\right) du

E=kλ0[ln⁡u+du]dd+LE = kλ_0 \left[\ln u + \frac{d}{u}\right]_d^{d+L}

E=kλ0[ln⁡(d+L)−ln⁡d+dd+L−1]E = kλ_0 \left[\ln(d+L) - \ln d + \frac{d}{d+L} - 1\right]

E=kλ0[ln⁡(d+Ld)+dd+L−1]E = kλ_0 \left[\ln\left(\frac{d+L}{d}\right) + \frac{d}{d+L} - 1\right]

Numerical calculation: E=(8.99×109)(4.0×10−6)[ln⁡(0.70.2)+0.20.7−1]E = (8.99 \times 10^9)(4.0 \times 10^{-6})\left[\ln\left(\frac{0.7}{0.2}\right) + \frac{0.2}{0.7} - 1\right]

E=35,960[ln⁡(3.5)+0.286−1]E = 35,960 \left[\ln(3.5) + 0.286 - 1\right]

E=35,960[1.253+0.286−1]=35,960(0.539)E = 35,960 \left[1.253 + 0.286 - 1\right] = 35,960(0.539)

E=1.94×104 N/CE = 1.94 \times 10^4 \text{ N/C}

Answer: The electric field at point P is 1.94 × 10⁴ N/C pointing away from the rod.

7Problem 7medium

❓ Question:

A ring of radius R = 0.1 m carries a uniformly distributed charge Q = 2.0 × 10⁻⁸ C. (a) Find the electric field at a point on the axis at distance x = 0.05 m from the center. (b) At what distance x does the electric field reach its maximum value? (c) What is the maximum field strength?

💡 Show Solution

Given:

  • R = 0.1 m
  • Q = 2.0 × 10⁻⁸ C = 20 nC
  • k = 8.99 × 10⁹ N·m²/C²

Formula for field on axis: Ex=kQx(x2+R2)3/2E_x = \frac{kQx}{(x^2 + R^2)^{3/2}}

(a) Field at x = 0.05 m:

E=(8.99×109)(2.0×10−8)(0.05)[(0.05)2+(0.1)2]3/2E = \frac{(8.99 \times 10^9)(2.0 \times 10^{-8})(0.05)}{[(0.05)^2 + (0.1)^2]^{3/2}}

E=(8.99)(0.1)[(0.0025+0.01)]3/2=0.899(0.0125)3/2E = \frac{(8.99)(0.1)}{[(0.0025 + 0.01)]^{3/2}} = \frac{0.899}{(0.0125)^{3/2}}

E=0.8991.40×10−3=6.42×102 N/CE = \frac{0.899}{1.40 \times 10^{-3}} = 6.42 \times 10^2 \text{ N/C}

(b) Maximum field position:

To find maximum, take derivative and set equal to zero: dExdx=kQddx[x(x2+R2)3/2]=0\frac{dE_x}{dx} = kQ\frac{d}{dx}\left[\frac{x}{(x^2 + R^2)^{3/2}}\right] = 0

dExdx=kQ(x2+R2)3/2−x⋅32(x2+R2)1/2(2x)(x2+R2)3\frac{dE_x}{dx} = kQ\frac{(x^2 + R^2)^{3/2} - x \cdot \frac{3}{2}(x^2 + R^2)^{1/2}(2x)}{(x^2 + R^2)^3}

=kQ(x2+R2)−3x2(x2+R2)5/2=0= kQ\frac{(x^2 + R^2) - 3x^2}{(x^2 + R^2)^{5/2}} = 0

R2−2x2=0R^2 - 2x^2 = 0

x=R2=0.12=0.0707 mx = \frac{R}{\sqrt{2}} = \frac{0.1}{\sqrt{2}} = 0.0707 \text{ m}

(c) Maximum field strength:

Emax=kQ(R/2)[(R/2)2+R2]3/2E_{max} = \frac{kQ(R/\sqrt{2})}{[(R/\sqrt{2})^2 + R^2]^{3/2}}

=kQ(R/2)[R2/2+R2]3/2=kQ(R/2)(3R2/2)3/2= \frac{kQ(R/\sqrt{2})}{[R^2/2 + R^2]^{3/2}} = \frac{kQ(R/\sqrt{2})}{(3R^2/2)^{3/2}}

=kQR22(3/2)3/2=kQR22⋅3322= \frac{kQ}{R^2\sqrt{2}(3/2)^{3/2}} = \frac{kQ}{R^2\sqrt{2} \cdot \frac{3\sqrt{3}}{2\sqrt{2}}}

=2kQ33R2=2(8.99×109)(2.0×10−8)33(0.1)2= \frac{2kQ}{3\sqrt{3}R^2} = \frac{2(8.99 \times 10^9)(2.0 \times 10^{-8})}{3\sqrt{3}(0.1)^2}

=359.60.052=6.91×102 N/C= \frac{359.6}{0.052} = 6.91 \times 10^2 \text{ N/C}

Answers: (a) E = 642 N/C (b) x = 7.07 cm = R/√2 (c) E_max = 691 N/C

8Problem 8medium

❓ Question:

A ring of radius R = 0.1 m carries a uniformly distributed charge Q = 2.0 × 10⁻⁸ C. (a) Find the electric field at a point on the axis at distance x = 0.05 m from the center. (b) At what distance x does the electric field reach its maximum value? (c) What is the maximum field strength?

💡 Show Solution

Given:

  • R = 0.1 m
  • Q = 2.0 × 10⁻⁸ C = 20 nC
  • k = 8.99 × 10⁹ N·m²/C²

Formula for field on axis: Ex=kQx(x2+R2)3/2E_x = \frac{kQx}{(x^2 + R^2)^{3/2}}

(a) Field at x = 0.05 m:

E=(8.99×109)(2.0×10−8)(0.05)[(0.05)2+(0.1)2]3/2E = \frac{(8.99 \times 10^9)(2.0 \times 10^{-8})(0.05)}{[(0.05)^2 + (0.1)^2]^{3/2}}

E=(8.99)(0.1)[(0.0025+0.01)]3/2=0.899(0.0125)3/2E = \frac{(8.99)(0.1)}{[(0.0025 + 0.01)]^{3/2}} = \frac{0.899}{(0.0125)^{3/2}}

E=0.8991.40×10−3=6.42×102 N/CE = \frac{0.899}{1.40 \times 10^{-3}} = 6.42 \times 10^2 \text{ N/C}

(b) Maximum field position:

To find maximum, take derivative and set equal to zero: dExdx=kQddx[x(x2+R2)3/2]=0\frac{dE_x}{dx} = kQ\frac{d}{dx}\left[\frac{x}{(x^2 + R^2)^{3/2}}\right] = 0

dExdx=kQ(x2+R2)3/2−x⋅32(x2+R2)1/2(2x)(x2+R2)3\frac{dE_x}{dx} = kQ\frac{(x^2 + R^2)^{3/2} - x \cdot \frac{3}{2}(x^2 + R^2)^{1/2}(2x)}{(x^2 + R^2)^3}

=kQ(x2+R2)−3x2(x2+R2)5/2=0= kQ\frac{(x^2 + R^2) - 3x^2}{(x^2 + R^2)^{5/2}} = 0

R2−2x2=0R^2 - 2x^2 = 0

x=R2=0.12=0.0707 mx = \frac{R}{\sqrt{2}} = \frac{0.1}{\sqrt{2}} = 0.0707 \text{ m}

(c) Maximum field strength:

Emax=kQ(R/2)[(R/2)2+R2]3/2E_{max} = \frac{kQ(R/\sqrt{2})}{[(R/\sqrt{2})^2 + R^2]^{3/2}}

=kQ(R/2)[R2/2+R2]3/2=kQ(R/2)(3R2/2)3/2= \frac{kQ(R/\sqrt{2})}{[R^2/2 + R^2]^{3/2}} = \frac{kQ(R/\sqrt{2})}{(3R^2/2)^{3/2}}

=kQR22(3/2)3/2=kQR22⋅3322= \frac{kQ}{R^2\sqrt{2}(3/2)^{3/2}} = \frac{kQ}{R^2\sqrt{2} \cdot \frac{3\sqrt{3}}{2\sqrt{2}}}

=2kQ33R2=2(8.99×109)(2.0×10−8)33(0.1)2= \frac{2kQ}{3\sqrt{3}R^2} = \frac{2(8.99 \times 10^9)(2.0 \times 10^{-8})}{3\sqrt{3}(0.1)^2}

=359.60.052=6.91×102 N/C= \frac{359.6}{0.052} = 6.91 \times 10^2 \text{ N/C}

Answers: (a) E = 642 N/C (b) x = 7.07 cm = R/√2 (c) E_max = 691 N/C

9Problem 9medium

❓ Question:

A ring of radius R = 0.1 m carries a uniformly distributed charge Q = 2.0 × 10⁻⁸ C. (a) Find the electric field at a point on the axis at distance x = 0.05 m from the center. (b) At what distance x does the electric field reach its maximum value? (c) What is the maximum field strength?

💡 Show Solution

Given:

  • R = 0.1 m
  • Q = 2.0 × 10⁻⁸ C = 20 nC
  • k = 8.99 × 10⁹ N·m²/C²

Formula for field on axis: Ex=kQx(x2+R2)3/2E_x = \frac{kQx}{(x^2 + R^2)^{3/2}}

(a) Field at x = 0.05 m:

E=(8.99×109)(2.0×10−8)(0.05)[(0.05)2+(0.1)2]3/2E = \frac{(8.99 \times 10^9)(2.0 \times 10^{-8})(0.05)}{[(0.05)^2 + (0.1)^2]^{3/2}}

E=(8.99)(0.1)[(0.0025+0.01)]3/2=0.899(0.0125)3/2E = \frac{(8.99)(0.1)}{[(0.0025 + 0.01)]^{3/2}} = \frac{0.899}{(0.0125)^{3/2}}

E=0.8991.40×10−3=6.42×102 N/CE = \frac{0.899}{1.40 \times 10^{-3}} = 6.42 \times 10^2 \text{ N/C}

(b) Maximum field position:

To find maximum, take derivative and set equal to zero: dExdx=kQddx[x(x2+R2)3/2]=0\frac{dE_x}{dx} = kQ\frac{d}{dx}\left[\frac{x}{(x^2 + R^2)^{3/2}}\right] = 0

dExdx=kQ(x2+R2)3/2−x⋅32(x2+R2)1/2(2x)(x2+R2)3\frac{dE_x}{dx} = kQ\frac{(x^2 + R^2)^{3/2} - x \cdot \frac{3}{2}(x^2 + R^2)^{1/2}(2x)}{(x^2 + R^2)^3}

=kQ(x2+R2)−3x2(x2+R2)5/2=0= kQ\frac{(x^2 + R^2) - 3x^2}{(x^2 + R^2)^{5/2}} = 0

R2−2x2=0R^2 - 2x^2 = 0

x=R2=0.12=0.0707 mx = \frac{R}{\sqrt{2}} = \frac{0.1}{\sqrt{2}} = 0.0707 \text{ m}

(c) Maximum field strength:

Emax=kQ(R/2)[(R/2)2+R2]3/2E_{max} = \frac{kQ(R/\sqrt{2})}{[(R/\sqrt{2})^2 + R^2]^{3/2}}

=kQ(R/2)[R2/2+R2]3/2=kQ(R/2)(3R2/2)3/2= \frac{kQ(R/\sqrt{2})}{[R^2/2 + R^2]^{3/2}} = \frac{kQ(R/\sqrt{2})}{(3R^2/2)^{3/2}}

=kQR22(3/2)3/2=kQR22⋅3322= \frac{kQ}{R^2\sqrt{2}(3/2)^{3/2}} = \frac{kQ}{R^2\sqrt{2} \cdot \frac{3\sqrt{3}}{2\sqrt{2}}}

=2kQ33R2=2(8.99×109)(2.0×10−8)33(0.1)2= \frac{2kQ}{3\sqrt{3}R^2} = \frac{2(8.99 \times 10^9)(2.0 \times 10^{-8})}{3\sqrt{3}(0.1)^2}

=359.60.052=6.91×102 N/C= \frac{359.6}{0.052} = 6.91 \times 10^2 \text{ N/C}

Answers: (a) E = 642 N/C (b) x = 7.07 cm = R/√2 (c) E_max = 691 N/C

10Problem 10medium

❓ Question:

A ring of radius R = 0.1 m carries a uniformly distributed charge Q = 2.0 × 10⁻⁸ C. (a) Find the electric field at a point on the axis at distance x = 0.05 m from the center. (b) At what distance x does the electric field reach its maximum value? (c) What is the maximum field strength?

💡 Show Solution

Given:

  • R = 0.1 m
  • Q = 2.0 × 10⁻⁸ C = 20 nC
  • k = 8.99 × 10⁹ N·m²/C²

Formula for field on axis: Ex=kQx(x2+R2)3/2E_x = \frac{kQx}{(x^2 + R^2)^{3/2}}

(a) Field at x = 0.05 m:

E=(8.99×109)(2.0×10−8)(0.05)[(0.05)2+(0.1)2]3/2E = \frac{(8.99 \times 10^9)(2.0 \times 10^{-8})(0.05)}{[(0.05)^2 + (0.1)^2]^{3/2}}

E=(8.99)(0.1)[(0.0025+0.01)]3/2=0.899(0.0125)3/2E = \frac{(8.99)(0.1)}{[(0.0025 + 0.01)]^{3/2}} = \frac{0.899}{(0.0125)^{3/2}}

E=0.8991.40×10−3=6.42×102 N/CE = \frac{0.899}{1.40 \times 10^{-3}} = 6.42 \times 10^2 \text{ N/C}

(b) Maximum field position:

To find maximum, take derivative and set equal to zero: dExdx=kQddx[x(x2+R2)3/2]=0\frac{dE_x}{dx} = kQ\frac{d}{dx}\left[\frac{x}{(x^2 + R^2)^{3/2}}\right] = 0

dExdx=kQ(x2+R2)3/2−x⋅32(x2+R2)1/2(2x)(x2+R2)3\frac{dE_x}{dx} = kQ\frac{(x^2 + R^2)^{3/2} - x \cdot \frac{3}{2}(x^2 + R^2)^{1/2}(2x)}{(x^2 + R^2)^3}

=kQ(x2+R2)−3x2(x2+R2)5/2=0= kQ\frac{(x^2 + R^2) - 3x^2}{(x^2 + R^2)^{5/2}} = 0

R2−2x2=0R^2 - 2x^2 = 0

x=R2=0.12=0.0707 mx = \frac{R}{\sqrt{2}} = \frac{0.1}{\sqrt{2}} = 0.0707 \text{ m}

(c) Maximum field strength:

Emax=kQ(R/2)[(R/2)2+R2]3/2E_{max} = \frac{kQ(R/\sqrt{2})}{[(R/\sqrt{2})^2 + R^2]^{3/2}}

=kQ(R/2)[R2/2+R2]3/2=kQ(R/2)(3R2/2)3/2= \frac{kQ(R/\sqrt{2})}{[R^2/2 + R^2]^{3/2}} = \frac{kQ(R/\sqrt{2})}{(3R^2/2)^{3/2}}

=kQR22(3/2)3/2=kQR22⋅3322= \frac{kQ}{R^2\sqrt{2}(3/2)^{3/2}} = \frac{kQ}{R^2\sqrt{2} \cdot \frac{3\sqrt{3}}{2\sqrt{2}}}

=2kQ33R2=2(8.99×109)(2.0×10−8)33(0.1)2= \frac{2kQ}{3\sqrt{3}R^2} = \frac{2(8.99 \times 10^9)(2.0 \times 10^{-8})}{3\sqrt{3}(0.1)^2}

=359.60.052=6.91×102 N/C= \frac{359.6}{0.052} = 6.91 \times 10^2 \text{ N/C}

Answers: (a) E = 642 N/C (b) x = 7.07 cm = R/√2 (c) E_max = 691 N/C

11Problem 11medium

❓ Question:

A ring of radius R = 0.1 m carries a uniformly distributed charge Q = 2.0 × 10⁻⁸ C. (a) Find the electric field at a point on the axis at distance x = 0.05 m from the center. (b) At what distance x does the electric field reach its maximum value? (c) What is the maximum field strength?

💡 Show Solution

Given:

  • R = 0.1 m
  • Q = 2.0 × 10⁻⁸ C = 20 nC
  • k = 8.99 × 10⁹ N·m²/C²

Formula for field on axis: Ex=kQx(x2+R2)3/2E_x = \frac{kQx}{(x^2 + R^2)^{3/2}}

(a) Field at x = 0.05 m:

E=(8.99×109)(2.0×10−8)(0.05)[(0.05)2+(0.1)2]3/2E = \frac{(8.99 \times 10^9)(2.0 \times 10^{-8})(0.05)}{[(0.05)^2 + (0.1)^2]^{3/2}}

E=(8.99)(0.1)[(0.0025+0.01)]3/2=0.899(0.0125)3/2E = \frac{(8.99)(0.1)}{[(0.0025 + 0.01)]^{3/2}} = \frac{0.899}{(0.0125)^{3/2}}

E=0.8991.40×10−3=6.42×102 N/CE = \frac{0.899}{1.40 \times 10^{-3}} = 6.42 \times 10^2 \text{ N/C}

(b) Maximum field position:

To find maximum, take derivative and set equal to zero: dExdx=kQddx[x(x2+R2)3/2]=0\frac{dE_x}{dx} = kQ\frac{d}{dx}\left[\frac{x}{(x^2 + R^2)^{3/2}}\right] = 0

dExdx=kQ(x2+R2)3/2−x⋅32(x2+R2)1/2(2x)(x2+R2)3\frac{dE_x}{dx} = kQ\frac{(x^2 + R^2)^{3/2} - x \cdot \frac{3}{2}(x^2 + R^2)^{1/2}(2x)}{(x^2 + R^2)^3}

=kQ(x2+R2)−3x2(x2+R2)5/2=0= kQ\frac{(x^2 + R^2) - 3x^2}{(x^2 + R^2)^{5/2}} = 0

R2−2x2=0R^2 - 2x^2 = 0

x=R2=0.12=0.0707 mx = \frac{R}{\sqrt{2}} = \frac{0.1}{\sqrt{2}} = 0.0707 \text{ m}

(c) Maximum field strength:

Emax=kQ(R/2)[(R/2)2+R2]3/2E_{max} = \frac{kQ(R/\sqrt{2})}{[(R/\sqrt{2})^2 + R^2]^{3/2}}

=kQ(R/2)[R2/2+R2]3/2=kQ(R/2)(3R2/2)3/2= \frac{kQ(R/\sqrt{2})}{[R^2/2 + R^2]^{3/2}} = \frac{kQ(R/\sqrt{2})}{(3R^2/2)^{3/2}}

=kQR22(3/2)3/2=kQR22⋅3322= \frac{kQ}{R^2\sqrt{2}(3/2)^{3/2}} = \frac{kQ}{R^2\sqrt{2} \cdot \frac{3\sqrt{3}}{2\sqrt{2}}}

=2kQ33R2=2(8.99×109)(2.0×10−8)33(0.1)2= \frac{2kQ}{3\sqrt{3}R^2} = \frac{2(8.99 \times 10^9)(2.0 \times 10^{-8})}{3\sqrt{3}(0.1)^2}

=359.60.052=6.91×102 N/C= \frac{359.6}{0.052} = 6.91 \times 10^2 \text{ N/C}

Answers: (a) E = 642 N/C (b) x = 7.07 cm = R/√2 (c) E_max = 691 N/C

12Problem 12medium

❓ Question:

A ring of radius R = 0.1 m carries a uniformly distributed charge Q = 2.0 × 10⁻⁸ C. (a) Find the electric field at a point on the axis at distance x = 0.05 m from the center. (b) At what distance x does the electric field reach its maximum value? (c) What is the maximum field strength?

💡 Show Solution

Given:

  • R = 0.1 m
  • Q = 2.0 × 10⁻⁸ C = 20 nC
  • k = 8.99 × 10⁹ N·m²/C²

Formula for field on axis: Ex=kQx(x2+R2)3/2E_x = \frac{kQx}{(x^2 + R^2)^{3/2}}

(a) Field at x = 0.05 m:

E=(8.99×109)(2.0×10−8)(0.05)[(0.05)2+(0.1)2]3/2E = \frac{(8.99 \times 10^9)(2.0 \times 10^{-8})(0.05)}{[(0.05)^2 + (0.1)^2]^{3/2}}

E=(8.99)(0.1)[(0.0025+0.01)]3/2=0.899(0.0125)3/2E = \frac{(8.99)(0.1)}{[(0.0025 + 0.01)]^{3/2}} = \frac{0.899}{(0.0125)^{3/2}}

E=0.8991.40×10−3=6.42×102 N/CE = \frac{0.899}{1.40 \times 10^{-3}} = 6.42 \times 10^2 \text{ N/C}

(b) Maximum field position:

To find maximum, take derivative and set equal to zero: dExdx=kQddx[x(x2+R2)3/2]=0\frac{dE_x}{dx} = kQ\frac{d}{dx}\left[\frac{x}{(x^2 + R^2)^{3/2}}\right] = 0

dExdx=kQ(x2+R2)3/2−x⋅32(x2+R2)1/2(2x)(x2+R2)3\frac{dE_x}{dx} = kQ\frac{(x^2 + R^2)^{3/2} - x \cdot \frac{3}{2}(x^2 + R^2)^{1/2}(2x)}{(x^2 + R^2)^3}

=kQ(x2+R2)−3x2(x2+R2)5/2=0= kQ\frac{(x^2 + R^2) - 3x^2}{(x^2 + R^2)^{5/2}} = 0

R2−2x2=0R^2 - 2x^2 = 0

x=R2=0.12=0.0707 mx = \frac{R}{\sqrt{2}} = \frac{0.1}{\sqrt{2}} = 0.0707 \text{ m}

(c) Maximum field strength:

Emax=kQ(R/2)[(R/2)2+R2]3/2E_{max} = \frac{kQ(R/\sqrt{2})}{[(R/\sqrt{2})^2 + R^2]^{3/2}}

=kQ(R/2)[R2/2+R2]3/2=kQ(R/2)(3R2/2)3/2= \frac{kQ(R/\sqrt{2})}{[R^2/2 + R^2]^{3/2}} = \frac{kQ(R/\sqrt{2})}{(3R^2/2)^{3/2}}

=kQR22(3/2)3/2=kQR22⋅3322= \frac{kQ}{R^2\sqrt{2}(3/2)^{3/2}} = \frac{kQ}{R^2\sqrt{2} \cdot \frac{3\sqrt{3}}{2\sqrt{2}}}

=2kQ33R2=2(8.99×109)(2.0×10−8)33(0.1)2= \frac{2kQ}{3\sqrt{3}R^2} = \frac{2(8.99 \times 10^9)(2.0 \times 10^{-8})}{3\sqrt{3}(0.1)^2}

=359.60.052=6.91×102 N/C= \frac{359.6}{0.052} = 6.91 \times 10^2 \text{ N/C}

Answers: (a) E = 642 N/C (b) x = 7.07 cm = R/√2 (c) E_max = 691 N/C

13Problem 13medium

❓ Question:

An electric dipole consists of charges +q and -q separated by distance d = 2.0 mm. The dipole moment is p = 5.0 × 10⁻¹² C·m. Find (a) the magnitude of each charge, (b) the electric field at a point on the perpendicular bisector at distance r = 10 cm (where r >> d), and (c) the electric field at a point on the axis at distance r = 10 cm from the center.

💡 Show Solution

Given:

  • d = 2.0 × 10⁻³ m = 2.0 mm
  • p = 5.0 × 10⁻¹² C·m
  • r = 0.1 m = 10 cm
  • k = 8.99 × 10⁹ N·m²/C²

(a) Magnitude of each charge:

Dipole moment: p = qd

q=pd=5.0×10−122.0×10−3=2.5×10−9 C=2.5 nCq = \frac{p}{d} = \frac{5.0 \times 10^{-12}}{2.0 \times 10^{-3}} = 2.5 \times 10^{-9} \text{ C} = 2.5 \text{ nC}

(b) Field on perpendicular bisector (far field, r >> d):

Eperp=kpr3E_{perp} = \frac{kp}{r^3}

E=(8.99×109)(5.0×10−12)(0.1)3E = \frac{(8.99 \times 10^9)(5.0 \times 10^{-12})}{(0.1)^3}

E=4.495×10−210−3=44.95 N/CE = \frac{4.495 \times 10^{-2}}{10^{-3}} = 44.95 \text{ N/C}

Direction: From +q toward -q (perpendicular to dipole axis)

(c) Field on axis (far field, r >> d):

Eaxis=2kpr3E_{axis} = \frac{2kp}{r^3}

E=2(8.99×109)(5.0×10−12)(0.1)3E = \frac{2(8.99 \times 10^9)(5.0 \times 10^{-12})}{(0.1)^3}

E=8.99×10−210−3=89.9 N/CE = \frac{8.99 \times 10^{-2}}{10^{-3}} = 89.9 \text{ N/C}

Direction: Away from dipole (from -q toward +q) for points on the side of +q

Note: The axial field is exactly twice the perpendicular field at the same distance: Eaxis=2EperpE_{axis} = 2E_{perp}

Answers: (a) q = 2.5 nC (b) E = 45.0 N/C (perpendicular to axis) (c) E = 89.9 N/C (along axis)**

14Problem 14medium

❓ Question:

An electric dipole consists of charges +q and -q separated by distance d = 2.0 mm. The dipole moment is p = 5.0 × 10⁻¹² C·m. Find (a) the magnitude of each charge, (b) the electric field at a point on the perpendicular bisector at distance r = 10 cm (where r >> d), and (c) the electric field at a point on the axis at distance r = 10 cm from the center.

💡 Show Solution

Given:

  • d = 2.0 × 10⁻³ m = 2.0 mm
  • p = 5.0 × 10⁻¹² C·m
  • r = 0.1 m = 10 cm
  • k = 8.99 × 10⁹ N·m²/C²

(a) Magnitude of each charge:

Dipole moment: p = qd

q=pd=5.0×10−122.0×10−3=2.5×10−9 C=2.5 nCq = \frac{p}{d} = \frac{5.0 \times 10^{-12}}{2.0 \times 10^{-3}} = 2.5 \times 10^{-9} \text{ C} = 2.5 \text{ nC}

(b) Field on perpendicular bisector (far field, r >> d):

Eperp=kpr3E_{perp} = \frac{kp}{r^3}

E=(8.99×109)(5.0×10−12)(0.1)3E = \frac{(8.99 \times 10^9)(5.0 \times 10^{-12})}{(0.1)^3}

E=4.495×10−210−3=44.95 N/CE = \frac{4.495 \times 10^{-2}}{10^{-3}} = 44.95 \text{ N/C}

Direction: From +q toward -q (perpendicular to dipole axis)

(c) Field on axis (far field, r >> d):

Eaxis=2kpr3E_{axis} = \frac{2kp}{r^3}

E=2(8.99×109)(5.0×10−12)(0.1)3E = \frac{2(8.99 \times 10^9)(5.0 \times 10^{-12})}{(0.1)^3}

E=8.99×10−210−3=89.9 N/CE = \frac{8.99 \times 10^{-2}}{10^{-3}} = 89.9 \text{ N/C}

Direction: Away from dipole (from -q toward +q) for points on the side of +q

Note: The axial field is exactly twice the perpendicular field at the same distance: Eaxis=2EperpE_{axis} = 2E_{perp}

Answers: (a) q = 2.5 nC (b) E = 45.0 N/C (perpendicular to axis) (c) E = 89.9 N/C (along axis)**

15Problem 15medium

❓ Question:

An electric dipole consists of charges +q and -q separated by distance d = 2.0 mm. The dipole moment is p = 5.0 × 10⁻¹² C·m. Find (a) the magnitude of each charge, (b) the electric field at a point on the perpendicular bisector at distance r = 10 cm (where r >> d), and (c) the electric field at a point on the axis at distance r = 10 cm from the center.

💡 Show Solution

Given:

  • d = 2.0 × 10⁻³ m = 2.0 mm
  • p = 5.0 × 10⁻¹² C·m
  • r = 0.1 m = 10 cm
  • k = 8.99 × 10⁹ N·m²/C²

(a) Magnitude of each charge:

Dipole moment: p = qd

q=pd=5.0×10−122.0×10−3=2.5×10−9 C=2.5 nCq = \frac{p}{d} = \frac{5.0 \times 10^{-12}}{2.0 \times 10^{-3}} = 2.5 \times 10^{-9} \text{ C} = 2.5 \text{ nC}

(b) Field on perpendicular bisector (far field, r >> d):

Eperp=kpr3E_{perp} = \frac{kp}{r^3}

E=(8.99×109)(5.0×10−12)(0.1)3E = \frac{(8.99 \times 10^9)(5.0 \times 10^{-12})}{(0.1)^3}

E=4.495×10−210−3=44.95 N/CE = \frac{4.495 \times 10^{-2}}{10^{-3}} = 44.95 \text{ N/C}

Direction: From +q toward -q (perpendicular to dipole axis)

(c) Field on axis (far field, r >> d):

Eaxis=2kpr3E_{axis} = \frac{2kp}{r^3}

E=2(8.99×109)(5.0×10−12)(0.1)3E = \frac{2(8.99 \times 10^9)(5.0 \times 10^{-12})}{(0.1)^3}

E=8.99×10−210−3=89.9 N/CE = \frac{8.99 \times 10^{-2}}{10^{-3}} = 89.9 \text{ N/C}

Direction: Away from dipole (from -q toward +q) for points on the side of +q

Note: The axial field is exactly twice the perpendicular field at the same distance: Eaxis=2EperpE_{axis} = 2E_{perp}

Answers: (a) q = 2.5 nC (b) E = 45.0 N/C (perpendicular to axis) (c) E = 89.9 N/C (along axis)**

16Problem 16medium

❓ Question:

An electric dipole consists of charges +q and -q separated by distance d = 2.0 mm. The dipole moment is p = 5.0 × 10⁻¹² C·m. Find (a) the magnitude of each charge, (b) the electric field at a point on the perpendicular bisector at distance r = 10 cm (where r >> d), and (c) the electric field at a point on the axis at distance r = 10 cm from the center.

💡 Show Solution

Given:

  • d = 2.0 × 10⁻³ m = 2.0 mm
  • p = 5.0 × 10⁻¹² C·m
  • r = 0.1 m = 10 cm
  • k = 8.99 × 10⁹ N·m²/C²

(a) Magnitude of each charge:

Dipole moment: p = qd

q=pd=5.0×10−122.0×10−3=2.5×10−9 C=2.5 nCq = \frac{p}{d} = \frac{5.0 \times 10^{-12}}{2.0 \times 10^{-3}} = 2.5 \times 10^{-9} \text{ C} = 2.5 \text{ nC}

(b) Field on perpendicular bisector (far field, r >> d):

Eperp=kpr3E_{perp} = \frac{kp}{r^3}

E=(8.99×109)(5.0×10−12)(0.1)3E = \frac{(8.99 \times 10^9)(5.0 \times 10^{-12})}{(0.1)^3}

E=4.495×10−210−3=44.95 N/CE = \frac{4.495 \times 10^{-2}}{10^{-3}} = 44.95 \text{ N/C}

Direction: From +q toward -q (perpendicular to dipole axis)

(c) Field on axis (far field, r >> d):

Eaxis=2kpr3E_{axis} = \frac{2kp}{r^3}

E=2(8.99×109)(5.0×10−12)(0.1)3E = \frac{2(8.99 \times 10^9)(5.0 \times 10^{-12})}{(0.1)^3}

E=8.99×10−210−3=89.9 N/CE = \frac{8.99 \times 10^{-2}}{10^{-3}} = 89.9 \text{ N/C}

Direction: Away from dipole (from -q toward +q) for points on the side of +q

Note: The axial field is exactly twice the perpendicular field at the same distance: Eaxis=2EperpE_{axis} = 2E_{perp}

Answers: (a) q = 2.5 nC (b) E = 45.0 N/C (perpendicular to axis) (c) E = 89.9 N/C (along axis)**

17Problem 17medium

❓ Question:

An electric dipole consists of charges +q and -q separated by distance d = 2.0 mm. The dipole moment is p = 5.0 × 10⁻¹² C·m. Find (a) the magnitude of each charge, (b) the electric field at a point on the perpendicular bisector at distance r = 10 cm (where r >> d), and (c) the electric field at a point on the axis at distance r = 10 cm from the center.

💡 Show Solution

Given:

  • d = 2.0 × 10⁻³ m = 2.0 mm
  • p = 5.0 × 10⁻¹² C·m
  • r = 0.1 m = 10 cm
  • k = 8.99 × 10⁹ N·m²/C²

(a) Magnitude of each charge:

Dipole moment: p = qd

q=pd=5.0×10−122.0×10−3=2.5×10−9 C=2.5 nCq = \frac{p}{d} = \frac{5.0 \times 10^{-12}}{2.0 \times 10^{-3}} = 2.5 \times 10^{-9} \text{ C} = 2.5 \text{ nC}

(b) Field on perpendicular bisector (far field, r >> d):

Eperp=kpr3E_{perp} = \frac{kp}{r^3}

E=(8.99×109)(5.0×10−12)(0.1)3E = \frac{(8.99 \times 10^9)(5.0 \times 10^{-12})}{(0.1)^3}

E=4.495×10−210−3=44.95 N/CE = \frac{4.495 \times 10^{-2}}{10^{-3}} = 44.95 \text{ N/C}

Direction: From +q toward -q (perpendicular to dipole axis)

(c) Field on axis (far field, r >> d):

Eaxis=2kpr3E_{axis} = \frac{2kp}{r^3}

E=2(8.99×109)(5.0×10−12)(0.1)3E = \frac{2(8.99 \times 10^9)(5.0 \times 10^{-12})}{(0.1)^3}

E=8.99×10−210−3=89.9 N/CE = \frac{8.99 \times 10^{-2}}{10^{-3}} = 89.9 \text{ N/C}

Direction: Away from dipole (from -q toward +q) for points on the side of +q

Note: The axial field is exactly twice the perpendicular field at the same distance: Eaxis=2EperpE_{axis} = 2E_{perp}

Answers: (a) q = 2.5 nC (b) E = 45.0 N/C (perpendicular to axis) (c) E = 89.9 N/C (along axis)**

18Problem 18medium

❓ Question:

An electric dipole consists of charges +q and -q separated by distance d = 2.0 mm. The dipole moment is p = 5.0 × 10⁻¹² C·m. Find (a) the magnitude of each charge, (b) the electric field at a point on the perpendicular bisector at distance r = 10 cm (where r >> d), and (c) the electric field at a point on the axis at distance r = 10 cm from the center.

💡 Show Solution

Given:

  • d = 2.0 × 10⁻³ m = 2.0 mm
  • p = 5.0 × 10⁻¹² C·m
  • r = 0.1 m = 10 cm
  • k = 8.99 × 10⁹ N·m²/C²

(a) Magnitude of each charge:

Dipole moment: p = qd

q=pd=5.0×10−122.0×10−3=2.5×10−9 C=2.5 nCq = \frac{p}{d} = \frac{5.0 \times 10^{-12}}{2.0 \times 10^{-3}} = 2.5 \times 10^{-9} \text{ C} = 2.5 \text{ nC}

(b) Field on perpendicular bisector (far field, r >> d):

Eperp=kpr3E_{perp} = \frac{kp}{r^3}

E=(8.99×109)(5.0×10−12)(0.1)3E = \frac{(8.99 \times 10^9)(5.0 \times 10^{-12})}{(0.1)^3}

E=4.495×10−210−3=44.95 N/CE = \frac{4.495 \times 10^{-2}}{10^{-3}} = 44.95 \text{ N/C}

Direction: From +q toward -q (perpendicular to dipole axis)

(c) Field on axis (far field, r >> d):

Eaxis=2kpr3E_{axis} = \frac{2kp}{r^3}

E=2(8.99×109)(5.0×10−12)(0.1)3E = \frac{2(8.99 \times 10^9)(5.0 \times 10^{-12})}{(0.1)^3}

E=8.99×10−210−3=89.9 N/CE = \frac{8.99 \times 10^{-2}}{10^{-3}} = 89.9 \text{ N/C}

Direction: Away from dipole (from -q toward +q) for points on the side of +q

Note: The axial field is exactly twice the perpendicular field at the same distance: Eaxis=2EperpE_{axis} = 2E_{perp}

Answers: (a) q = 2.5 nC (b) E = 45.0 N/C (perpendicular to axis) (c) E = 89.9 N/C (along axis)**

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