Physics · Electrostatics · Practice set
Electrostatics Practice · Level 1 and Level 2 with Solutions
Twenty graded Electrostatics questions for JEE: ten Level 1 to lock in the basics, ten Level 2 on superposition, dipoles and Gauss's law, each with a full worked solution.
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20Questions
L1 + L2Levels
FullSolutions
In short
Electrostatics questions separate into two levels. Level 1 asks you to apply Coulomb's law, field and potential directly. Level 2 asks you to superpose several contributions, handle a dipole, or pick the right Gaussian surface. This set works through ten of each, with the full solution shown every time.
Contents
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Concept Recap
The core results these questions test, at a glance. Level 1 covers Coulomb's law, field and potential; Level 2 adds superposition, dipoles, capacitance and Gauss's law.
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Electrostatics – 20 MCQs (L1 + L2)
Single correct answer. The first ten are Level 1 foundations; the last ten are Level 2. Each is tagged with its level and topic.
Q1LEVEL 1Coulomb's Law
Two point charges of +2 μC and +3 μC are placed 0.3 m apart in vacuum. The force between them is:
A0.3 N
B0.06 N
C6 N
D0.6 N
Q2LEVEL 1Electric Field
The electric field at 0.1 m from a point charge of 1 μC is:
A9 × 10⁷ N/C
B9 × 10⁴ N/C
C9 × 10⁵ N/C
D9 × 10⁶ N/C
Q3LEVEL 1Potential
The electric potential at 0.1 m from a point charge of 1 μC is:
A9 × 10⁶ V
B9 × 10³ V
C9 × 10⁴ V
D9 × 10⁵ V
Q4LEVEL 1Field Lines
Electric field lines around a single positive point charge:
APoint radially outward
BAre parallel straight lines
CPoint radially inward
DForm closed loops
Q5LEVEL 1Electric Flux
The SI unit of electric flux is:
AV/m
BC/m²
CN/C
DN m²/C
Q6LEVEL 1Coulomb's Law
Two equal charges q are separated by distance r. If r is doubled, the force becomes:
AOne half
BOne quarter
CDouble
DFour times
Q7LEVEL 1Potential Energy
The electric potential energy of two charges +2 μC and +3 μC separated by 0.3 m is:
A0.18 J
B0.06 J
C1.8 J
D0.6 J
Q8LEVEL 1Conductors
Inside a hollow charged conductor (no charge enclosed), the electric field is:
ARadially outward
BEqual to the surface field
CZero
DUniform and non-zero
Q9LEVEL 1Dipole
The dipole moment of two charges ±q separated by 2a points:
APerpendicular to the axis
BRadially outward
CFrom +q to −q
DFrom −q to +q
Q10LEVEL 1Quantisation
One coulomb of charge equals the charge of about:
A1.6 × 10⁻¹⁹ electrons
B6.25 × 10¹⁸ electrons
C9 × 10⁹ electrons
D6.25 × 10²³ electrons
Q11LEVEL 2Superposition
Three equal charges q sit at the corners of an equilateral triangle of side a. The net force on any one charge is:
Akq²/a²
B2 kq²/a²
C3 kq²/a²
D√3 kq²/a²
Q12LEVEL 2Null Point
A charge +q and a charge +4q are separated by distance d. The point where the net electric field is zero lies:
AAt d/3 from +q
BAt d/3 from +4q
CAt 2d/3 from +q
DAt d/2 from +q
Q13LEVEL 2Dipole
The electric field on the axis of a short dipole at distance r is E₁; on the equatorial line at the same r it is E₂. The ratio E₁:E₂ is:
A2 : 1
B1 : 2
C1 : 1
D4 : 1
Q14LEVEL 2Dipole in Field
A dipole of moment p is placed in a uniform field E at angle θ. The torque on it is:
ApE tanθ
BpE sinθ
CpE
DpE cosθ
Q15LEVEL 2Equipotential
The work done in moving a charge between two points on the same equipotential surface is:
APositive
BZero
CNegative
DqV
Q16LEVEL 2Capacitance
A parallel-plate capacitor has capacitance C. If a dielectric of constant K fills the gap, the new capacitance is:
AC + K
BKC
CC/(1+K)
DC/K
Q17LEVEL 2Energy
The energy stored in a capacitor of capacitance C charged to voltage V is:
ACV²
B2CV²
C½CV
D½CV²
Q18LEVEL 2Conductors
The electric field just outside a charged conductor of surface charge density σ is:
A2σ/ε₀
Bσ/2ε₀
Cσ/ε₀
Dzero
Q19LEVEL 2Charge Sharing
Two identical charged spheres attract with force F. They are touched together and returned to the same separation. If the charges were +3q and −q, the new force is:
AAttractive, F/3
BAttractive, F
CRepulsive, F/3
DRepulsive, F
Q20LEVEL 2Gauss's Law
A charge q is placed at the centre of a cube. The electric flux through one face is:
Aq/6ε₀
B6q/ε₀
Cq/ε₀
Dq/8ε₀
★
Answer Key & Solutions
Q1D
Q2C
Q3C
Q4A
Q5D
Q6B
Q7A
Q8C
Q9D
Q10B
Q11D
Q12A
Q13A
Q14B
Q15B
Q16B
Q17D
Q18C
Q19C
Q20A
Q1Correct: D0.6 N
F = kq₁q₂/r² = (9×10⁹)(2×10⁻⁶)(3×10⁻⁶)/(0.3)² = (9×10⁹)(6×10⁻¹²)/0.09 = 0.6 N. The charges are alike, so the force is repulsive.
Q2Correct: C9 × 10⁵ N/C
E = kq/r² = (9×10⁹)(10⁻⁶)/(0.1)² = (9×10³)/(0.01) = 9×10⁵ N/C. Field falls off as 1/r².
Q3Correct: C9 × 10⁴ V
V = kq/r = (9×10⁹)(10⁻⁶)/0.1 = 9×10⁴ V. Potential falls off as 1/r, more slowly than the field.
Q4Correct: APoint radially outward
Field lines start on positive charge and point radially outward. They point inward for a negative charge and never form closed loops in electrostatics.
Q5Correct: DN m²/C
Electric flux φ = E · A has units of (N/C)(m²) = N m²/C, equivalent to V m.
Q6Correct: BOne quarter
F ∝ 1/r², so doubling r gives F₂ = F/2² = F/4. The inverse-square law makes the force drop fast with distance.
Q7Correct: A0.18 J
U = kq₁q₂/r = (9×10⁹)(6×10⁻¹²)/0.3 = 0.18 J. Positive, since both charges are positive (work was needed to bring them together).
Q8Correct: CZero
By Gauss's law, with no enclosed charge the field inside a conductor's cavity is zero. All excess charge resides on the outer surface.
Q9Correct: DFrom −q to +q
By convention the dipole moment p = q(2a) points from the negative to the positive charge, along the axis.
Q10Correct: B6.25 × 10¹⁸ electrons
Number = 1/e = 1/(1.6×10⁻¹⁹) ≈ 6.25×10¹⁸ electrons. Charge is quantised in units of e.
Q11Correct: D√3 kq²/a²
Each of the two other charges exerts F = kq²/a² at 60° to each other. Resultant = √(F²+F²+2F²cos60°) = F√3 = √3 kq²/a², directed away from the triangle's centre.
Q12Correct: AAt d/3 from +q
Set kq/x² = k(4q)/(d−x)². Then (d−x)/x = 2, so d−x = 2x, giving x = d/3 from +q. The null point is nearer the smaller charge.
Q13Correct: A2 : 1
Axial field = 2kp/r³, equatorial field = kp/r³, so E₁:E₂ = 2 : 1. The axial field is always twice the equatorial for the same distance.
Q14Correct: BpE sinθ
Torque τ = p × E = pE sinθ, maximum at θ = 90° and zero when the dipole aligns with the field (θ = 0).
Q15Correct: BZero
Both points are at the same potential, so ΔV = 0 and W = qΔV = zero. No work is needed along an equipotential.
Q16Correct: BKC
A dielectric increases capacitance by the factor K: C' = KC. The dielectric reduces the field for a given charge, lowering the voltage and raising C = Q/V.
Q17Correct: D½CV²
Energy U = ½CV² = ½QV = Q²/2C. It equals the work done charging the capacitor against the rising voltage.
Q18Correct: Cσ/ε₀
Just outside a conductor the field is σ/ε₀, twice that of an isolated sheet (σ/2ε₀), because all the field is pushed into the outer half-space.
Q19Correct: CRepulsive, F/3
On touching, charge equalises to (3q − q)/2 = +q on each. New force ∝ (q)(q) = q², while original ∝ (3q)(q) = 3q², so new = F/3 and now repulsive (both positive).
Q20Correct: Aq/6ε₀
Total flux through the cube = q/ε₀ (Gauss). By symmetry each of the six faces gets an equal share: q/6ε₀.
★
Electrostatics: Quick Revision
Everything for revision day.COULOMB
F = kq₁q₂/r²
k = 9×10⁹
F ∝ 1/r²
like: repel, unlike: attract
FIELD / POTENTIAL
E = kq/r²
V = kq/r
E = −dV/dr
field: 1/r², V: 1/r
DIPOLE
p = q(2a), − to +
axial E = 2kp/r³
equatorial E = kp/r³
torque = pE sinθ
GAUSS
φ = q/ε₀
sheet: σ/2ε₀
conductor: σ/ε₀
inside conductor: E=0
CAPACITOR
C = ε₀A/d
dielectric: C→KC
U = ½CV²
E = V/d
KEY TRICKS
null pt of q,4q: d/3 from q
triangle net: √3 kq²/a²
equipotential: W=0
cube face: q/6ε₀
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