1
Electrostatics Mastery: 200 Multiple-Choice
Questions with Answers and Explanations
1. Two identical small conducting spheres carry charges +4 μC and −2 μC, and are touched
together then separated. What is the final charge on each?
● A) +4 μC and −2 μC
● B) +1 μC and +1 μC
● C) +2 μC and 0 μC
● D) +3 μC and −1 μC
Answer: B. Explanation: Total charge = +2 μC; identical conductors share equally, so
each gets +1 μC.
2. The SI unit of charge is:
● A) Volt
● B) Coulomb
● C) Ampere
● D) Farad
Answer: B. Explanation: Coulomb (C) is the SI unit of electric charge.
3. The charge of an electron is approximately:
● A) +1.6 × 10^−19 C
● B) −1.6 × 10^−19 C
● C) −9.1 × 10^−31 C
● D) +9.1 × 10^−31 C
Answer: B. Explanation: Electron has negative elementary charge magnitude e ≈ 1.602 ×
10^−19 C.
4. Coulomb’s law in vacuum states the force magnitude between two point charges q1 and
q2 separated by r is:
● A) k q1 q2 / r
● B) k q1 q2 / r^2
● C) k q1 q2 r^2
● D) k q1 q2 r
Answer: B. Explanation: Inverse-square dependence with k = 1/(4π ε0).
5. The constant 1/(4π ε0) numerically equals approximately:
● A) 8.99 × 10^9 N·m^2/C^2
● B) 9.81 m/s^2
● C) 1.60 × 10^−19 C
, 2
● D) 3.00 × 10^8 m/s
Answer: A. Explanation: k ≈ 8.99 × 10^9 N·m^2/C^2.
6. If the medium has relative permittivity εr, Coulomb’s law uses:
● A) k/εr
● B) k εr
● C) ε0 εr
● D) 1/(4π ε0 εr)
Answer: D. Explanation: Effective constant k’ = 1/(4π ε0 εr).
7. Electric field due to point charge q at distance r is:
● A) E = k q r^2
● B) E = k q / r^2
● C) E = k q / r
● D) E = k q r
Answer: B. Explanation: Field magnitude decays as inverse square.
8. Direction of electric field due to a positive point charge is:
● A) Tangential to circles around charge
● B) Radially inward
● C) Radially outward
● D) Undefined
Answer: C. Explanation: Field lines point away from positive charge.
9. The principle of superposition applies to electrostatics because:
● A) Fields are vectors and Maxwell’s equations are linear in charge
● B) Fields are scalars
● C) Charges are quantized
● D) Energy is conserved
Answer: A. Explanation: Linear relation E ∝ ρ enables vector addition.
10.A test charge is:
● A) Always positive and large
● B) Infinitesimal so it doesn’t disturb the field
● C) Negative to probe positive charges
● D) A neutral object
Answer: B. Explanation: Ideal test charge does not alter source distribution.
11.The electric field inside a conductor in electrostatic equilibrium is:
● A) Nonzero and uniform
, 3
● B) Zero
● C) Infinite
● D) Radial
Answer: B. Explanation: Free charges rearrange to cancel internal fields.
12.The charge resides in a conductor at electrostatic equilibrium:
● A) Uniformly in the bulk
● B) Only on the surface
● C) Only at the center
● D) Only at edges
Answer: B. Explanation: Internal E = 0 implies charges on the surface.
13.For a conductor, the electric field just outside the surface is:
● A) Tangential
● B) Zero
● C) Normal with magnitude σ/ε0
● D) Normal with magnitude ε0/σ
Answer: C. Explanation: Boundary condition E⊥ = σ/ε0 (in vacuum).
14.Electric flux through a closed surface S is:
● A) ∮ E · dl
● B) ∮ B · dA
● C) ∮ E · dA
● D) ∮ V · dA
Answer: C. Explanation: Flux is surface integral of E normal component.
15.Gauss’s law in integral form:
● A) ∮ E · dA = 0
● B) ∮ E · dA = Q_enc/ε0
● C) ∮ B · dA = μ0 I
● D) ∮ E · dl = −dΦB/dt
Answer: B. Explanation: Net flux equals enclosed charge over ε0.
16.A spherical Gaussian surface of radius r around a point charge q gives E equal to:
● A) k q / r
● B) k q / r^2
● C) q/(4π r^2)
● D) q/(4π ε0 r^2)
Answer: D. Explanation: From Gauss’s law: E (4π r^2) = q/ε0.
17.An infinite line charge with linear density λ has E at distance r:
, 4
● A) E = λ/(2π ε0 r)
● B) E = k λ/r^2
● C) E = λ/(4π ε0 r^2)
● D) E = 2k λ/r
Answer: A. Explanation: From cylindrical symmetry and Gauss’s law.
18.An infinite plane with surface charge density σ has E:
● A) σ/(4ε0) on each side
● B) σ/ε0 on each side
● C) σ/(2ε0) on each side
● D) 2σ/ε0 on each side
Answer: C. Explanation: Magnitude σ/(2ε0) on each side, pointing away if σ > 0.
19.For two oppositely charged infinite planes ±σ, the field between them is:
● A) Zero
● B) σ/ε0
● C) σ/(2ε0)
● D) 2σ/ε0
Answer: B. Explanation: Fields add between plates, cancel outside.
20.Electric potential difference between two points is:
● A) Path-dependent always
● B) Negative integral of E · dl
● C) Magnetic flux
● D) Charge per unit area
Answer: B. Explanation: V(b) − V(a) = −∫_a^b E · dl in electrostatics.
21.Potential due to point charge q at distance r (reference at infinity):
● A) V = k q r
● B) V = k q / r
● C) V = k r / q
● D) V = q/(4π r)
Answer: B. Explanation: V = 1/(4π ε0) q/r.
22.The relation between E and V:
● A) E = ∇ × V
● B) E = −∇V
● C) V = ∇ · E
● D) ∇ · E = 0
Answer: B. Explanation: Electrostatic field is conservative: curl E = 0.
Electrostatics Mastery: 200 Multiple-Choice
Questions with Answers and Explanations
1. Two identical small conducting spheres carry charges +4 μC and −2 μC, and are touched
together then separated. What is the final charge on each?
● A) +4 μC and −2 μC
● B) +1 μC and +1 μC
● C) +2 μC and 0 μC
● D) +3 μC and −1 μC
Answer: B. Explanation: Total charge = +2 μC; identical conductors share equally, so
each gets +1 μC.
2. The SI unit of charge is:
● A) Volt
● B) Coulomb
● C) Ampere
● D) Farad
Answer: B. Explanation: Coulomb (C) is the SI unit of electric charge.
3. The charge of an electron is approximately:
● A) +1.6 × 10^−19 C
● B) −1.6 × 10^−19 C
● C) −9.1 × 10^−31 C
● D) +9.1 × 10^−31 C
Answer: B. Explanation: Electron has negative elementary charge magnitude e ≈ 1.602 ×
10^−19 C.
4. Coulomb’s law in vacuum states the force magnitude between two point charges q1 and
q2 separated by r is:
● A) k q1 q2 / r
● B) k q1 q2 / r^2
● C) k q1 q2 r^2
● D) k q1 q2 r
Answer: B. Explanation: Inverse-square dependence with k = 1/(4π ε0).
5. The constant 1/(4π ε0) numerically equals approximately:
● A) 8.99 × 10^9 N·m^2/C^2
● B) 9.81 m/s^2
● C) 1.60 × 10^−19 C
, 2
● D) 3.00 × 10^8 m/s
Answer: A. Explanation: k ≈ 8.99 × 10^9 N·m^2/C^2.
6. If the medium has relative permittivity εr, Coulomb’s law uses:
● A) k/εr
● B) k εr
● C) ε0 εr
● D) 1/(4π ε0 εr)
Answer: D. Explanation: Effective constant k’ = 1/(4π ε0 εr).
7. Electric field due to point charge q at distance r is:
● A) E = k q r^2
● B) E = k q / r^2
● C) E = k q / r
● D) E = k q r
Answer: B. Explanation: Field magnitude decays as inverse square.
8. Direction of electric field due to a positive point charge is:
● A) Tangential to circles around charge
● B) Radially inward
● C) Radially outward
● D) Undefined
Answer: C. Explanation: Field lines point away from positive charge.
9. The principle of superposition applies to electrostatics because:
● A) Fields are vectors and Maxwell’s equations are linear in charge
● B) Fields are scalars
● C) Charges are quantized
● D) Energy is conserved
Answer: A. Explanation: Linear relation E ∝ ρ enables vector addition.
10.A test charge is:
● A) Always positive and large
● B) Infinitesimal so it doesn’t disturb the field
● C) Negative to probe positive charges
● D) A neutral object
Answer: B. Explanation: Ideal test charge does not alter source distribution.
11.The electric field inside a conductor in electrostatic equilibrium is:
● A) Nonzero and uniform
, 3
● B) Zero
● C) Infinite
● D) Radial
Answer: B. Explanation: Free charges rearrange to cancel internal fields.
12.The charge resides in a conductor at electrostatic equilibrium:
● A) Uniformly in the bulk
● B) Only on the surface
● C) Only at the center
● D) Only at edges
Answer: B. Explanation: Internal E = 0 implies charges on the surface.
13.For a conductor, the electric field just outside the surface is:
● A) Tangential
● B) Zero
● C) Normal with magnitude σ/ε0
● D) Normal with magnitude ε0/σ
Answer: C. Explanation: Boundary condition E⊥ = σ/ε0 (in vacuum).
14.Electric flux through a closed surface S is:
● A) ∮ E · dl
● B) ∮ B · dA
● C) ∮ E · dA
● D) ∮ V · dA
Answer: C. Explanation: Flux is surface integral of E normal component.
15.Gauss’s law in integral form:
● A) ∮ E · dA = 0
● B) ∮ E · dA = Q_enc/ε0
● C) ∮ B · dA = μ0 I
● D) ∮ E · dl = −dΦB/dt
Answer: B. Explanation: Net flux equals enclosed charge over ε0.
16.A spherical Gaussian surface of radius r around a point charge q gives E equal to:
● A) k q / r
● B) k q / r^2
● C) q/(4π r^2)
● D) q/(4π ε0 r^2)
Answer: D. Explanation: From Gauss’s law: E (4π r^2) = q/ε0.
17.An infinite line charge with linear density λ has E at distance r:
, 4
● A) E = λ/(2π ε0 r)
● B) E = k λ/r^2
● C) E = λ/(4π ε0 r^2)
● D) E = 2k λ/r
Answer: A. Explanation: From cylindrical symmetry and Gauss’s law.
18.An infinite plane with surface charge density σ has E:
● A) σ/(4ε0) on each side
● B) σ/ε0 on each side
● C) σ/(2ε0) on each side
● D) 2σ/ε0 on each side
Answer: C. Explanation: Magnitude σ/(2ε0) on each side, pointing away if σ > 0.
19.For two oppositely charged infinite planes ±σ, the field between them is:
● A) Zero
● B) σ/ε0
● C) σ/(2ε0)
● D) 2σ/ε0
Answer: B. Explanation: Fields add between plates, cancel outside.
20.Electric potential difference between two points is:
● A) Path-dependent always
● B) Negative integral of E · dl
● C) Magnetic flux
● D) Charge per unit area
Answer: B. Explanation: V(b) − V(a) = −∫_a^b E · dl in electrostatics.
21.Potential due to point charge q at distance r (reference at infinity):
● A) V = k q r
● B) V = k q / r
● C) V = k r / q
● D) V = q/(4π r)
Answer: B. Explanation: V = 1/(4π ε0) q/r.
22.The relation between E and V:
● A) E = ∇ × V
● B) E = −∇V
● C) V = ∇ · E
● D) ∇ · E = 0
Answer: B. Explanation: Electrostatic field is conservative: curl E = 0.
