Electric Fields and Electric Potential

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

⚡ 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.

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

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:

Lines start on positive charges, end on negative charges
Density of lines ∝ field strength
Lines never cross
Tangent to line gives field direction
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:

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

For Potential:

Calculate $V$ from each charge (scalar!)
Algebraic sum (watch signs)
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)

📚 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: $q = +2.0 \times 10^{-6}$ C
Distance: $r = 0.30$ m
Test charge: $q_0 = -3.0 \times 10^{-6}$ C

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

2Problem 2easy

❓ 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: $q = +2.0 \times 10^{-6}$ C
Distance: $r = 0.30$ m
Test charge: $q_0 = -3.0 \times 10^{-6}$ C

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

3Problem 3medium

❓ Question:

A point charge of +4.0 μC is located at the origin. (a) What is the electric field at a point 0.50 m away? (b) What is the electric potential at this point? (c) How much work is required to bring a +2.0 μC charge from infinity to this point? Use k = 9.0 × 10⁹ N·m²/C².

💡 Show Solution

Solution:

Given: Q = +4.0 × 10⁻⁶ C, r = 0.50 m, k = 9.0 × 10⁹ N·m²/C²

(a) Electric field: E = kQ/r² = (9.0 × 10⁹)(4.0 × 10⁻⁶)/(0.50)² E = (36 × 10³)/0.25 E = 1.44 × 10⁵ N/C or 144 kN/C (radially outward)

(b) Electric potential: V = kQ/r = (9.0 × 10⁹)(4.0 × 10⁻⁶)/0.50 V = (36 × 10³)/0.50 V = 7.2 × 10⁴ V or 72 kV

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

4Problem 4medium

❓ Question:

💡 Show Solution

Solution:

Given: Q = +4.0 × 10⁻⁶ C, r = 0.50 m, k = 9.0 × 10⁹ N·m²/C²

(a) Electric field: E = kQ/r² = (9.0 × 10⁹)(4.0 × 10⁻⁶)/(0.50)² E = (36 × 10³)/0.25 E = 1.44 × 10⁵ N/C or 144 kN/C (radially outward)

(b) Electric potential: V = kQ/r = (9.0 × 10⁹)(4.0 × 10⁻⁶)/0.50 V = (36 × 10³)/0.50 V = 7.2 × 10⁴ V or 72 kV

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

5Problem 5medium

❓ 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: $d = 2.0$ cm $= 0.020$ m
Potential difference: $\Delta V = 100$ V
Electron charge: $e = -1.60 \times 10^{-19}$ C

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

6Problem 6medium

❓ 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: $d = 2.0$ cm $= 0.020$ m
Potential difference: $\Delta V = 100$ V
Electron charge: $e = -1.60 \times 10^{-19}$ C

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

7Problem 7hard

❓ Question:

Two parallel plates separated by 0.020 m have a potential difference of 100 V. (a) What is the electric field between the plates? (b) What force acts on an electron in this field? (c) If an electron starts from rest at the negative plate, what is its speed when it reaches the positive plate? Use e = 1.6 × 10⁻¹⁹ C, m_e = 9.1 × 10⁻³¹ kg.

💡 Show Solution

Solution:

Given: d = 0.020 m, V = 100 V, e = 1.6 × 10⁻¹⁹ C, m = 9.1 × 10⁻³¹ kg

(a) Electric field: E = V/d = 100/0.020 = 5000 V/m or 5.0 × 10³ N/C (uniform field between parallel plates)

(b) Force on electron: F = eE = (1.6 × 10⁻¹⁹)(5000) F = 8.0 × 10⁻¹⁶ N (toward positive plate)

(c) Final speed: Using energy: ΔKE = Work = eV ½mv² - 0 = eV v² = 2eV/m = 2(1.6 × 10⁻¹⁹)(100)/(9.1 × 10⁻³¹) v² = (32 × 10⁻¹⁸)/(9.1 × 10⁻³¹) = 3.52 × 10¹³ v = 5.9 × 10⁶ m/s

8Problem 8hard

❓ 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:

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

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

Distance from $q_1$ : $r_1 = 0.20$ 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)

9Problem 9hard

❓ Question:

💡 Show Solution

Given:

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

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

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

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)

10Problem 10hard

❓ Question:

💡 Show Solution

Solution:

Given: d = 0.020 m, V = 100 V, e = 1.6 × 10⁻¹⁹ C, m = 9.1 × 10⁻³¹ kg

(a) Electric field: E = V/d = 100/0.020 = 5000 V/m or 5.0 × 10³ N/C (uniform field between parallel plates)

(b) Force on electron: F = eE = (1.6 × 10⁻¹⁹)(5000) F = 8.0 × 10⁻¹⁶ N (toward positive plate)

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