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

where:

• $\vec{E}$ = electric field (N/C or V/m)
• $\vec{F}$ = force on test charge
• $q_0$ = small positive test charge

Point Charge Field:

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

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

$\vec{E}_{total} = \vec{E}_1 + \vec{E}_2 + \vec{E}_3 + ...$

Calculate field from each charge, then vector sum.

Electric Potential Energy

Work done moving charge in electric field:

$U_E = k\frac{q_1 q_2}{r}$

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

Change in PE: $\Delta U = -W_{field} = W_{external}$

Electric Potential (Voltage)

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

$V = \frac{U}{q_0} = k\frac{q}{r}$

Units: Volt (V) = J/C

Potential Difference:

$\Delta V = V_f - V_i = -\int \vec{E} \cdot d\vec{r}$

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

where $d$ is distance in field direction.

Relationship: E and V

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

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

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

Parallel Plate Capacitor

Uniform field between plates:

$E = \frac{V}{d} = \frac{\sigma}{\epsilon_0}$

where:

• $V$ = potential difference
• $d$ = plate separation
• $\sigma$ = surface charge density

Equipotential Surfaces

Surfaces where $V =$ 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:

$1 \text{ eV} = 1.60 \times 10^{-19} \text{ J}$

Useful for atomic/particle physics.

Problem-Solving Strategy

For Fields:

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

For Potential:

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

Common Mistakes

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

Part (a): Electric field

$E = k\frac{|q|}{r^2} = (9.0 \times 10^9)\frac{2.0 \times 10^{-6}}{(0.30)^2}$ $E = (9.0 \times 10^9)\frac{2.0 \times 10^{-6}}{0.09} = 2.0 \times 10^5 \text{ N/C}$

Direction: Away from positive charge

Part (b): Force on test charge

$F = |q_0|E = (3.0 \times 10^{-6})(2.0 \times 10^5) = 0.60 \text{ N}$

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

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

Part (a): Electric field

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

Part (b): Force on electron

$F = |e|E = (1.60 \times 10^{-19})(5000)$ $F = 8.0 \times 10^{-16} \text{ N}$

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

Part (a): Electric field

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

Part (b): Force on electron

$F = |e|E = (1.60 \times 10^{-19})(5000)$ $F = 8.0 \times 10^{-16} \text{ N}$

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

Positive work (energy input required) to bring positive charge toward positive charge.

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

Distance from $q_1$: $r_1 = 0.20$ m Distance from $q_2$: $r_2 = 0.40 - 0.20 = 0.20$ m

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

Part (b): Work to bring charge from infinity

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

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

Answer:

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

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

Distance from $q_1$: $r_1 = 0.20$ m Distance from $q_2$: $r_2 = 0.40 - 0.20 = 0.20$ m

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

Part (b): Work to bring charge from infinity

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

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

Answer:

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