Electric field vectors, field lines, electric potential energy, and voltage
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Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 8 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
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Electric Fields and Electric Potential is part of the AP Physics 2 course on Study Mondo, specifically in the Electrostatics section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Electric Fields and Electric Potential?
q0
F
where:
E = electric field (N/C or V/m)
F = force on test charge
q0 = small positive test charge
Point Charge Field:
E=kr2∣q∣
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
Electric potential (V) is potential energy per unit charge:
V=q0U=krq
Units: Volt (V) = J/C
Potential Difference:
ΔV=Vf−Vi=−∫E⋅dr
For uniform field:
ΔV=−Ed
where d is distance in field direction.
Relationship: E and V
E=−drdV
Electric field points from high to low potential (downhill).
For uniform field:
E=dΔV
Parallel Plate Capacitor
Uniform field between plates:
E=dV=ϵ0σ
where:
V = potential difference
d = plate separation
σ = 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 eV=1.60×10−19 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 ΔV=−Ed for uniform field
Common Mistakes
❌ Treating potential as vector (it's scalar!)
❌ Wrong field direction from negative charge
❌ Forgetting ΔV=Vf−Vi (order matters)
❌ Sign errors in potential energy
❌ Confusing E (field) with V (potential)
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=+2.0×10−6 C
Distance: r=0.30 m
Test charge: q0=−3.0×10−6 C
Part (a): Electric field
E=kr2∣q∣
Direction: Away from positive charge
Part (b): Force on test charge
F=∣q0∣E=(3.0×10
Direction: Toward source (opposite to field, negative charge)
Answer: E = 2.0 × 10⁵ N/C, F = 0.60 N toward source
3Problem 3medium
❓ 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: ΔV=100 V
Electron charge: e=−1.60×10−19 C
Part (a): Electric field
E=dΔV=0.020
Part (b): Force on electron
F=∣e∣E=(1.60×10−19)(5000)
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
4Problem 4medium
❓ 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: ΔV=100 V
Electron charge: e=−1.60×10−19 C
Part (a): Electric field
E=dΔV=0.020
Part (b): Force on electron
F=∣e∣E=(1.60×10−19)(5000)
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
5Problem 5medium
❓ 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
(c) Work required:
W = qΔV = q(V_f - V_∞)
W = (2.0 × 10⁻⁶)(7.2 × 10⁴ - 0)
W = 0.144 J or 144 mJ
Positive work (energy input required) to bring positive charge toward positive charge.
6Problem 6hard
❓ 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=+5.0×10−6 C at x = 0
q2=−3.0×10−6 C at x = 0.40 m
Point of interest: x = 0.20 m
Part (a): Electric potential at x = 0.20 m
Distance from q1: r1=0.20 m
Distance from : m
Potential is scalar, so algebraic sum:
V=V1+V2=
Part (b): Work to bring charge from infinity
At infinity: Vi=0
At x = 0.20 m: Vf=90 kV
Work by external force:
W=qΔV=q(Vf−Vi
Answer:
(a) V = 90 kV
(b) W = 0.18 J (positive, work must be done against field)
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:
q1=+5.0×10−6 C at x = 0
q2=−3.0×10−6 C at x = 0.40 m
Point of interest: x = 0.20 m
Part (a): Electric potential at x = 0.20 m
Distance from q1: r1=0.20 m
Distance from : m
Potential is scalar, so algebraic sum:
V=V1+V2=
Part (b): Work to bring charge from infinity
At infinity: Vi=0
At x = 0.20 m: Vf=90 kV
Work by external force:
W=qΔV=q(Vf−Vi
Answer:
(a) V = 90 kV
(b) W = 0.18 J (positive, work must be done against field)
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Yes, this page includes 8 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.