Electrostatics
Q. In electrostatics, what is the significance of equipotential surfaces?
A.
They are surfaces where electric field is zero
B.
They are surfaces where potential is constant
C.
They are surfaces where charge density is uniform
D.
They are surfaces where current flows
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Solution
Equipotential surfaces are defined as surfaces where the electric potential is constant. No work is done when moving a charge along these surfaces.
Correct Answer: B — They are surfaces where potential is constant
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Q. The electric potential due to a uniformly charged sphere at a point outside the sphere is equivalent to that of?
A.
A point charge at the center
B.
A point charge at the surface
C.
A point charge at the edge
D.
A hollow sphere
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Solution
The electric potential outside a uniformly charged sphere is the same as that due to a point charge located at the center of the sphere.
Correct Answer: A — A point charge at the center
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Q. The potential energy of a system of two point charges q1 and q2 separated by a distance r is given by?
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. The work done in moving a charge from a point A to point B in an electric field is equal to the change in what?
A.
Electric potential energy
B.
Electric potential
C.
Electric field strength
D.
Charge
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Solution
The work done in moving a charge in an electric field is equal to the change in electric potential energy.
Correct Answer: A — Electric potential energy
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Q. Two capacitors, C1 = 2μF and C2 = 3μF, are connected in series. What is the equivalent capacitance?
A.
1.2μF
B.
5μF
C.
6μF
D.
0.6μF
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Solution
For capacitors in series, the equivalent capacitance C_eq is given by 1/C_eq = 1/C1 + 1/C2. Thus, 1/C_eq = 1/2 + 1/3 = 5/6, so C_eq = 6/5 = 1.2μF.
Correct Answer: A — 1.2μF
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Q. Two capacitors, C1 and C2, are connected in series. What is the equivalent capacitance?
A.
C1 + C2
B.
1 / (1/C1 + 1/C2)
C.
C1 * C2 / (C1 + C2)
D.
C1 - C2
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Solution
The equivalent capacitance of capacitors in series is given by 1 / (1/C1 + 1/C2).
Correct Answer: B — 1 / (1/C1 + 1/C2)
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Q. Two charges +q and -q are placed at a distance d apart. What is the electric field at the midpoint?
A.
0
B.
k * q / (d/2)^2
C.
k * q / d^2
D.
k * q / (d^2) * 2
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Solution
At the midpoint, the electric fields due to both charges cancel each other out, resulting in a net electric field of 0.
Correct Answer: A — 0
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Q. Two charges +q and -q are placed at a distance d apart. What is the electric field at the midpoint between the charges?
A.
0
B.
kq/d^2
C.
2kq/d^2
D.
kq/2d^2
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Solution
The electric fields due to both charges at the midpoint cancel each other, resulting in a net electric field of 0.
Correct Answer: A — 0
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Q. Two charges +q and -q are placed at a distance d apart. Where can a third charge be placed such that the net force on it is zero?
A.
At a distance d/2 from +q
B.
At a distance d/2 from -q
C.
At a distance greater than d from both
D.
At a distance less than d/2 from both
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Solution
The third charge must be placed between +q and -q, closer to -q to balance the forces.
Correct Answer: B — At a distance d/2 from -q
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Q. Two charges +q and -q are placed at a distance d apart. Where is the electric field zero?
A.
At the midpoint
B.
Closer to +q
C.
Closer to -q
D.
At infinity
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Solution
The electric field is zero at a point closer to -q because the magnitudes of the fields due to both charges will be equal at that point.
Correct Answer: C — Closer to -q
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Q. Two identical charges are placed 1m apart. If the force between them is 9N, what is the magnitude of each charge?
A.
1μC
B.
2μC
C.
3μC
D.
4μC
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Solution
Using Coulomb's law, F = k * |q1 * q2| / r². Let q be the charge, then 9 = (9 × 10^9) * (q²) / (1)². Solving gives q = 3μC.
Correct Answer: B — 2μC
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Q. Two identical charges of +1μC are placed 1m apart. What is the potential energy of the system?
A.
9 × 10^-3 J
B.
4.5 × 10^-3 J
C.
1.8 × 10^-3 J
D.
0.9 × 10^-3 J
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Solution
U = k * q1 * q2 / r = (9 × 10^9) * (1 × 10^-6) * (1 × 10^-6) / 1 = 9 × 10^-3 J.
Correct Answer: A — 9 × 10^-3 J
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Q. Two identical metal spheres carry charges of +5μC and -5μC respectively. If they are brought into contact and then separated, what will be the charge on each sphere?
A.
0μC
B.
+5μC
C.
-5μC
D.
+2.5μC
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Solution
When brought into contact, the charges will redistribute equally, resulting in 0μC on each sphere.
Correct Answer: A — 0μC
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Q. Two identical metal spheres carry charges of +5μC and -5μC. If they are brought into contact and then separated, what will be the charge on each sphere?
A.
0μC
B.
+5μC
C.
-5μC
D.
+2.5μC
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Solution
When brought into contact, the charges will redistribute equally, resulting in 0μC on each sphere.
Correct Answer: A — 0μC
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Q. Two identical spheres, each with a charge of +5μC, are placed 0.1m apart. What is the electric field at the midpoint between the two spheres?
A.
0 N/C
B.
1.8 × 10^5 N/C
C.
3.6 × 10^5 N/C
D.
9 × 10^5 N/C
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Solution
The electric fields due to both charges at the midpoint are equal and opposite, thus they cancel each other out, resulting in 0 N/C.
Correct Answer: A — 0 N/C
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Q. Two identical spheres, each with a charge of +5μC, are placed 1 meter apart. What is the potential energy of the system?
A.
0.225 J
B.
0.45 J
C.
0.675 J
D.
0.9 J
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Solution
Potential energy U = k * q1 * q2 / r = (9 × 10^9 N m²/C²) * (5 × 10^-6 C)² / 1 m = 0.225 J.
Correct Answer: B — 0.45 J
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Q. Two identical spheres, one charged positively and the other negatively, are brought into contact and then separated. What will be the charge on each sphere after separation?
A.
Both positive
B.
Both negative
C.
Neutral
D.
Equal positive and negative
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Solution
When two identical spheres are brought into contact, they share their charges equally. Thus, they will have equal positive and negative charges after separation.
Correct Answer: D — Equal positive and negative
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Q. Two point charges of +3 μC and -3 μC are placed 1 m apart. What is the electric potential at the midpoint?
A.
0 V
B.
9 × 10^9 V
C.
4.5 × 10^9 V
D.
None of the above
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Solution
The potentials due to both charges at the midpoint cancel each other, so V = 0 V.
Correct Answer: A — 0 V
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Q. Two point charges, +q and -q, are placed a distance d apart. What is the electric field at the midpoint between the charges?
A.
0
B.
k * q / (d/2)^2
C.
k * q / d^2
D.
k * q / (d^2) * 2
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Solution
At the midpoint, the electric fields due to both charges cancel each other out, resulting in a net electric field of 0.
Correct Answer: A — 0
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Q. Two point charges, +Q and -Q, are placed at a distance d apart. What is the electric potential at the midpoint between them?
A.
0
B.
kQ/d
C.
kQ/2d
D.
kQ/4d
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Solution
At the midpoint, the potentials due to both charges cancel each other out, resulting in a net potential of 0 V.
Correct Answer: A — 0
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Q. Two point charges, +Q and -Q, are placed at a distance d apart. What is the electric potential at the midpoint between the charges?
A.
0
B.
kQ/d
C.
kQ/2d
D.
kQ/d²
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Solution
The electric potential at the midpoint is zero because the potentials due to +Q and -Q cancel each other out.
Correct Answer: A — 0
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Q. Using Gauss's law, what is the electric field inside a uniformly charged cylindrical shell of radius R?
A.
0
B.
Q/(2πε₀R)
C.
Q/(4πε₀R²)
D.
Q/(2πε₀R²)
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Solution
Inside a uniformly charged cylindrical shell, the electric field is zero due to symmetry, as the contributions from all parts of the shell cancel out.
Correct Answer: A — 0
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Q. What does Gauss's law relate to in electrostatics?
A.
Electric field and charge distribution
B.
Magnetic field and current
C.
Pressure and volume
D.
Temperature and heat
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Solution
Gauss's law states that the electric flux through a closed surface is proportional to the charge enclosed within that surface.
Correct Answer: A — Electric field and charge distribution
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Q. What happens to the capacitance of a capacitor if the dielectric constant is doubled?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
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Solution
The capacitance C of a capacitor is directly proportional to the dielectric constant k. If k is doubled, C also doubles.
Correct Answer: A — It doubles
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Q. What happens to the capacitance of a parallel plate capacitor if the distance between the plates is doubled?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
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Solution
Capacitance C = ε₀A/d; if d is doubled, C is halved.
Correct Answer: B — It halves
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Q. What happens to the capacitance of a parallel plate capacitor if the distance between the plates is halved?
A.
It doubles
B.
It halves
C.
It quadruples
D.
It remains the same
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Solution
Capacitance, C = ε₀ * A / d. If d is halved, C doubles.
Correct Answer: A — It doubles
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Q. What happens to the electric field if the charge is tripled while keeping the distance constant?
A.
It triples
B.
It halves
C.
It remains the same
D.
It quadruples
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Solution
The electric field is directly proportional to the charge. Tripling the charge will triple the electric field.
Correct Answer: A — It triples
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Q. What happens to the electric field inside a conductor in electrostatic equilibrium?
A.
It is zero
B.
It is constant
C.
It varies linearly
D.
It is maximum at the center
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Solution
In electrostatic equilibrium, the electric field inside a conductor is zero.
Correct Answer: A — It is zero
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Q. What happens to the electric field inside a conductor when it is in electrostatic equilibrium?
A.
It increases
B.
It decreases
C.
It becomes zero
D.
It remains constant
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Solution
In electrostatic equilibrium, the electric field inside a conductor is zero.
Correct Answer: C — It becomes zero
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Q. What happens to the electric field inside a conductor when it reaches electrostatic equilibrium?
A.
It becomes uniform
B.
It becomes zero
C.
It increases
D.
It decreases
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Solution
In electrostatic equilibrium, the electric field inside a conductor is zero.
Correct Answer: B — It becomes zero
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