Electric Fields and Electric Potential

Electric field vectors, field lines, electric potential energy, and voltage

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⚡ Electric Fields and Electric Potential

Electric Field

An electric field is a vector field that describes the electric force per unit charge at any point in space.

E=Fq0\vec{E} = \frac{\vec{F}}{q_0}

where:

  • E\vec{E} = electric field (N/C or V/m)
  • F\vec{F} = force on test charge
  • q0q_0 = small positive test charge

Point Charge Field:

E=kqr2E = k\frac{|q|}{r^2}

Direction:

  • Positive charge → field points away
  • Negative charge → field points toward

Electric Field Lines

Visual representation of electric fields:

Rules:

  1. Lines start on positive charges, end on negative charges
  2. Density of lines ∝ field strength
  3. Lines never cross
  4. Tangent to line gives field direction
  5. Perpendicular to conductor surfaces

Uniform field: Parallel, evenly spaced lines (e.g., parallel plates)


Superposition of Fields

Etotal=E1+E2+E3+...\vec{E}_{total} = \vec{E}_1 + \vec{E}_2 + \vec{E}_3 + ...

Calculate field from each charge, then vector sum.


Electric Potential Energy

Work done moving charge in electric field:

UE=kq1q2rU_E = k\frac{q_1 q_2}{r}

  • Same sign charges: U>0U > 0 (repulsive, stored energy)
  • Opposite sign charges: U<0U < 0 (attractive, bound state)

Change in PE: ΔU=Wfield=Wexternal\Delta U = -W_{field} = W_{external}


Electric Potential (Voltage)

Electric potential (V) is potential energy per unit charge:

V=Uq0=kqrV = \frac{U}{q_0} = k\frac{q}{r}

Units: Volt (V) = J/C

Potential Difference:

ΔV=VfVi=Edr\Delta V = V_f - V_i = -\int \vec{E} \cdot d\vec{r}

For uniform field: ΔV=Ed\Delta V = -Ed

where dd is distance in field direction.


Relationship: E and V

E=dVdr\vec{E} = -\frac{dV}{dr}

Electric field points from high to low potential (downhill).

For uniform field: E=ΔVdE = \frac{\Delta V}{d}


Parallel Plate Capacitor

Uniform field between plates:

E=Vd=σϵ0E = \frac{V}{d} = \frac{\sigma}{\epsilon_0}

where:

  • VV = potential difference
  • dd = plate separation
  • σ\sigma = surface charge density

Equipotential Surfaces

Surfaces where V=V = constant

  • No work to move charge along equipotential
  • Always ⊥ to electric field lines
  • Closer spacing → stronger field

Electron Volt (eV)

Energy gained by electron moving through 1 V:

1 eV=1.60×1019 J1 \text{ eV} = 1.60 \times 10^{-19} \text{ J}

Useful for atomic/particle physics.


Problem-Solving Strategy

For Fields:

  1. Calculate EE from each charge
  2. Determine directions (away from +, toward -)
  3. Use components if needed
  4. Vector sum

For Potential:

  1. Calculate VV from each charge (scalar!)
  2. Algebraic sum (watch signs)
  3. Or use ΔV=Ed\Delta V = -Ed for uniform field

Common Mistakes

❌ Treating potential as vector (it's scalar!) ❌ Wrong field direction from negative charge ❌ Forgetting ΔV=VfVi\Delta V = V_f - V_i (order matters) ❌ Sign errors in potential energy ❌ Confusing E (field) with V (potential)

📚 Practice Problems

1Problem 1easy

Question:

A +2.0 μC charge creates an electric field. What is the field strength 0.30 m away? What force would a -3.0 μC charge experience at that point?

💡 Show Solution

Given:

  • Source: q=+2.0×106q = +2.0 \times 10^{-6} C
  • Distance: r=0.30r = 0.30 m
  • Test charge: q0=3.0×106q_0 = -3.0 \times 10^{-6} C

Part (a): Electric field

E=kqr2=(9.0×109)2.0×106(0.30)2E = k\frac{|q|}{r^2} = (9.0 \times 10^9)\frac{2.0 \times 10^{-6}}{(0.30)^2} E=(9.0×109)2.0×1060.09=2.0×105 N/CE = (9.0 \times 10^9)\frac{2.0 \times 10^{-6}}{0.09} = 2.0 \times 10^5 \text{ N/C}

Direction: Away from positive charge

Part (b): Force on test charge

F=q0E=(3.0×106)(2.0×105)=0.60 NF = |q_0|E = (3.0 \times 10^{-6})(2.0 \times 10^5) = 0.60 \text{ N}

Direction: Toward source (opposite to field, negative charge)

Answer: E = 2.0 × 10⁵ N/C, F = 0.60 N toward source

2Problem 2medium

Question:

Two parallel plates are 2.0 cm apart with a potential difference of 100 V. (a) What is the electric field between the plates? (b) What force acts on an electron between the plates?

💡 Show Solution

Given:

  • Plate separation: d=2.0d = 2.0 cm =0.020= 0.020 m
  • Potential difference: ΔV=100\Delta V = 100 V
  • Electron charge: e=1.60×1019e = -1.60 \times 10^{-19} C

Part (a): Electric field

E=ΔVd=1000.020=5000 N/C=5.0 kN/CE = \frac{\Delta V}{d} = \frac{100}{0.020} = 5000 \text{ N/C} = 5.0 \text{ kN/C}

Part (b): Force on electron

F=eE=(1.60×1019)(5000)F = |e|E = (1.60 \times 10^{-19})(5000) F=8.0×1016 NF = 8.0 \times 10^{-16} \text{ N}

Direction: Toward positive plate (opposite to field direction)

Answer:

  • (a) E = 5.0 kN/C
  • (b) F = 8.0 × 10⁻¹⁶ N toward positive plate

3Problem 3hard

Question:

A +5.0 μC charge is at the origin and a -3.0 μC charge is at x = 0.40 m. (a) Find the electric potential at x = 0.20 m. (b) How much work is required to bring a +2.0 μC charge from infinity to x = 0.20 m?

💡 Show Solution

Given:

  • q1=+5.0×106q_1 = +5.0 \times 10^{-6} C at x = 0
  • q2=3.0×106q_2 = -3.0 \times 10^{-6} C at x = 0.40 m
  • Point of interest: x = 0.20 m

Part (a): Electric potential at x = 0.20 m

Distance from q1q_1: r1=0.20r_1 = 0.20 m Distance from q2q_2: r2=0.400.20=0.20r_2 = 0.40 - 0.20 = 0.20 m

Potential is scalar, so algebraic sum: V=V1+V2=kq1r1+kq2r2V = V_1 + V_2 = k\frac{q_1}{r_1} + k\frac{q_2}{r_2} V=(9.0×109)[5.0×1060.20+3.0×1060.20]V = (9.0 \times 10^9)\left[\frac{5.0 \times 10^{-6}}{0.20} + \frac{-3.0 \times 10^{-6}}{0.20}\right] V=(9.0×109)[2.0×1060.20]V = (9.0 \times 10^9)\left[\frac{2.0 \times 10^{-6}}{0.20}\right] V=(9.0×109)(1.0×105)V = (9.0 \times 10^9)(1.0 \times 10^{-5}) V=9.0×104 V=90 kVV = 9.0 \times 10^4 \text{ V} = 90 \text{ kV}

Part (b): Work to bring charge from infinity

At infinity: Vi=0V_i = 0 At x = 0.20 m: Vf=90V_f = 90 kV

Work by external force: W=qΔV=q(VfVi)W = q\Delta V = q(V_f - V_i) W=(2.0×106)(9.0×1040)W = (2.0 \times 10^{-6})(9.0 \times 10^4 - 0) W=0.18 JW = 0.18 \text{ J}

Answer:

  • (a) V = 90 kV
  • (b) W = 0.18 J (positive, work must be done against field)