Electrostatics
Q. What is the potential difference between two points in an electric field if the work done to move a charge of 2μC is 0.04 J?
A.
20 V
B.
30 V
C.
40 V
D.
50 V
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Solution
Potential difference V = W/q = 0.04 J / (2 × 10^-6 C) = 20,000 V = 20 V.
Correct Answer: C — 40 V
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Q. What is the potential difference between two points in an electric field if the work done in moving a charge of 3 C from one point to another is 15 J?
A.
5 V
B.
10 V
C.
15 V
D.
20 V
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Solution
Potential difference V = Work done / Charge = 15 J / 3 C = 5 V.
Correct Answer: B — 10 V
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Q. What is the potential difference between two points in an electric field if the work done in moving a charge of +2μC between them is 0.04 J?
A.
20 V
B.
10 V
C.
5 V
D.
2 V
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Solution
V = W/q = 0.04 J / (2 × 10^-6 C) = 20 V.
Correct Answer: A — 20 V
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Q. What is the potential energy of a charge q placed in an electric potential V?
A.
qV
B.
q/V
C.
V/q
D.
q²V
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Solution
The potential energy U of a charge q in an electric potential V is given by U = qV.
Correct Answer: A — qV
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Q. What is the potential energy of a dipole in a uniform electric field E when it is aligned with the field?
A.
-pE
B.
0
C.
pE
D.
pE/2
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Solution
Potential energy U = -p · E. When aligned, U = -pE.
Correct Answer: A — -pE
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Q. What is the potential energy of a system of two charges +3μC and +4μC separated by 0.2m?
A.
0.54 J
B.
0.72 J
C.
0.36 J
D.
0.18 J
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Solution
U = k * (q1 * q2) / r = (9 × 10^9) * (3 × 10^-6 * 4 × 10^-6) / 0.2 = 0.54 J.
Correct Answer: B — 0.72 J
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Q. What is the potential energy of a system of two charges +3μC and +5μC separated by 0.2m?
A.
-6.75 J
B.
6.75 J
C.
0.75 J
D.
-0.75 J
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Solution
Potential energy U = k * (q1 * q2) / r = (9 × 10^9) * (3 × 10^-6) * (5 × 10^-6) / 0.2 = 6.75 J.
Correct Answer: B — 6.75 J
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Q. What is the potential energy of a system of two charges +3μC and -5μC separated by 0.3m?
A.
-0.45 J
B.
0.45 J
C.
-0.15 J
D.
0.15 J
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Solution
Potential energy U = k * q1 * q2 / r = (9 × 10^9) * (3 × 10^-6) * (-5 × 10^-6) / 0.3 = -0.45 J.
Correct Answer: A — -0.45 J
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Q. What is the potential energy of a system of two charges +5μC and +10μC separated by 0.2m?
A.
-0.225 J
B.
0.225 J
C.
0.45 J
D.
0.9 J
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Solution
Potential energy U = k * (q1 * q2) / r = (9 × 10^9) * (5 × 10^-6) * (10 × 10^-6) / 0.2 = 0.225 J.
Correct Answer: B — 0.225 J
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Q. What is the potential energy of a system of two charges +q and +q separated by a distance r?
A.
0
B.
k*q²/r
C.
-k*q²/r
D.
2k*q²/r
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Solution
Potential energy U = k * (q1 * q2) / r = k * (q * q) / r = k*q²/r.
Correct Answer: B — k*q²/r
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Q. What is the potential energy of a system of two charges of +1μC and -1μC separated by 0.1m?
A.
-0.09 J
B.
0.09 J
C.
-0.18 J
D.
0.18 J
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Solution
Potential energy U = k * q1 * q2 / r = (9 × 10^9) * (1 × 10^-6) * (-1 × 10^-6) / 0.1 = -0.09 J.
Correct Answer: A — -0.09 J
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Q. What is the potential energy of a system of two point charges +q and -q separated by a distance r?
A.
0
B.
kq²/r
C.
-kq²/r
D.
kq/r
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Solution
The potential energy U of a system of two point charges is given by U = k(q1q2)/r = -kq²/r for +q and -q.
Correct Answer: C — -kq²/r
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Q. What is the potential energy of a system of two point charges q1 and q2 separated by a distance r?
A.
k * q1 * q2 / r
B.
k * q1 * q2 * r
C.
k * (q1 + q2) / r
D.
k * (q1 - q2) / r
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Solution
The potential energy U of a system of two point charges is given by U = k * q1 * q2 / r.
Correct Answer: A — k * q1 * q2 / r
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Q. What is the relationship between electric field and electric potential?
A.
E = -dV/dr
B.
E = dV/dr
C.
E = V/r
D.
E = V^2
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Solution
The electric field E is the negative gradient of the electric potential V, given by E = -dV/dr.
Correct Answer: A — E = -dV/dr
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Q. What is the relationship between electric field E and electric potential V?
A.
E = -dV/dx
B.
E = dV/dx
C.
E = V/d
D.
E = -V/d
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Solution
The electric field E is the negative gradient of the electric potential V, given by E = -dV/dx.
Correct Answer: A — E = -dV/dx
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Q. What is the relationship between electric field E and potential V in a uniform electric field?
A.
E = -dV/dx
B.
E = dV/dx
C.
E = V/dx
D.
E = -V/dx
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Solution
In a uniform electric field, the relationship is given by E = -dV/dx, indicating that the electric field points in the direction of decreasing potential.
Correct Answer: A — E = -dV/dx
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Q. What is the relationship between electric field E and potential V in electrostatics?
A.
E = -dV/dx
B.
E = dV/dx
C.
E = V/dx
D.
E = -V
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Solution
The electric field is the negative gradient of the electric potential, E = -dV/dx.
Correct Answer: A — E = -dV/dx
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Q. What is the relationship between electric field E and surface charge density σ on a conductor?
A.
E = σ/ε₀
B.
E = σ/2ε₀
C.
E = 2σ/ε₀
D.
E = σ/4ε₀
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Solution
The electric field just outside a charged conductor is given by E = σ/ε₀.
Correct Answer: A — E = σ/ε₀
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Q. What is the relationship between electric potential (V) and electric field (E) in a uniform field?
A.
V = E × d
B.
V = E/d
C.
E = V × d
D.
E = V/d
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Solution
In a uniform electric field, the electric potential difference V is given by V = E × d, where d is the distance moved in the direction of the field.
Correct Answer: A — V = E × d
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Q. What is the relationship between electric potential and electric field?
A.
E = -dV/dx
B.
E = dV/dx
C.
E = V/d
D.
E = -V/d
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Solution
The electric field E is the negative gradient of the electric potential V, expressed as E = -dV/dx.
Correct Answer: A — E = -dV/dx
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Q. What is the total electric flux through a closed surface surrounding a charge of -3Q?
A.
-3Q/ε₀
B.
3Q/ε₀
C.
0
D.
-Q/ε₀
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Solution
The total electric flux through a closed surface is given by Φ = Q_enc/ε₀, where Q_enc is the enclosed charge.
Correct Answer: A — -3Q/ε₀
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Q. What is the unit of capacitance?
A.
Farad
B.
Coulomb
C.
Volt
D.
Ohm
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Solution
The unit of capacitance is the Farad (F), which is defined as one coulomb per volt.
Correct Answer: A — Farad
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Q. What is the unit of electric field?
A.
Volt
B.
Newton/Coulomb
C.
Coulomb
D.
Joule
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Solution
The electric field is defined as the force per unit charge, hence its unit is Newton per Coulomb (N/C).
Correct Answer: B — Newton/Coulomb
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Q. What is the unit of electric potential?
A.
Coulomb
B.
Volt
C.
Joule
D.
Newton
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Solution
The unit of electric potential is the Volt (V), which is defined as one Joule per Coulomb.
Correct Answer: B — Volt
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Q. What is the work done in moving a charge of +2μC from a point A to B in a uniform electric field of 500 N/C over a distance of 0.4m?
A.
400 J
B.
200 J
C.
100 J
D.
80 J
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Solution
Work done W = F * d = q * E * d = (2 × 10^-6 C) * (500 N/C) * (0.4 m) = 0.4 J = 80 J.
Correct Answer: D — 80 J
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Q. What is the work done in moving a charge of +2μC through a potential difference of 5V?
A.
10 μJ
B.
5 μJ
C.
2 μJ
D.
1 μJ
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Solution
Work done W = q * V = (2 × 10^-6 C) * (5 V) = 10 μJ.
Correct Answer: A — 10 μJ
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Q. What is the work done in moving a charge of +2μC through a potential difference of 10V?
A.
20 μJ
B.
200 μJ
C.
2 μJ
D.
0.2 μJ
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Solution
W = q * V = 2 × 10^-6 C * 10 V = 20 × 10^-6 J = 20 μJ.
Correct Answer: B — 200 μJ
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Q. What is the work done in moving a charge of 2 μC from a point at 50 V to a point at 100 V?
A.
0.1 mJ
B.
0.2 mJ
C.
0.3 mJ
D.
0.4 mJ
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Solution
Work done W = q * ΔV = 2 × 10^-6 C * (100 V - 50 V) = 2 × 10^-6 * 50 = 0.1 mJ.
Correct Answer: B — 0.2 mJ
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Q. What is the work done in moving a charge of 2 μC from a point at 50 V to another at 100 V?
A.
100 μJ
B.
200 μJ
C.
150 μJ
D.
50 μJ
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Solution
Work done W = q * ΔV = (2 × 10^-6 C) * (100 V - 50 V) = 100 μJ.
Correct Answer: B — 200 μJ
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Q. Which of the following statements about electric field lines is true?
A.
They can cross each other
B.
They are always straight
C.
They point from positive to negative charges
D.
They are always circular
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Solution
Electric field lines point from positive to negative charges, indicating the direction of the electric field.
Correct Answer: C — They point from positive to negative charges
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