Electrostatics
Q. If two capacitors of capacitance 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 C_eq of capacitors in series is given by 1 / C_eq = 1 / C1 + 1 / C2.
Correct Answer: B — 1 / (1/C1 + 1/C2)
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Q. If two charges of +3μC and +5μC are placed 0.3m apart, what is the magnitude of the force between them?
A.
0.15 N
B.
0.25 N
C.
0.45 N
D.
0.75 N
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Solution
Using Coulomb's law, F = k * |q1 * q2| / r² = (9 × 10^9 N m²/C²) * |(3 × 10^-6 C) * (5 × 10^-6 C)| / (0.3 m)² = 0.45 N.
Correct Answer: C — 0.45 N
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Q. If two charges of +3μC and -3μC are placed 0.1m apart, what is the net electric field at the midpoint?
A.
0 N/C
B.
54000 N/C
C.
27000 N/C
D.
81000 N/C
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Solution
The electric fields due to both charges at the midpoint cancel each other out, resulting in a net electric field of 0 N/C.
Correct Answer: A — 0 N/C
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Q. If two identical charges are brought closer together, what happens to the potential energy of the system?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes zero
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Solution
As the charges are brought closer, the potential energy of the system increases due to the repulsive force between like charges.
Correct Answer: A — Increases
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Q. If two identical charges are placed 1 meter apart, what happens to the force between them if the distance is doubled?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
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Solution
According to Coulomb's law, force is inversely proportional to the square of the distance. If the distance is doubled, the force becomes one-fourth.
Correct Answer: B — It halves
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Q. If two identical charges are placed 1m apart, what is the electric field at a point midway between them?
A.
0 N/C
B.
kq/2
C.
kq
D.
2kq
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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 N/C
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Q. If two identical charges are placed at a distance of 1m apart, what is the electric field at a point midway between them?
A.
0 N/C
B.
kq/2
C.
kq
D.
kq/4
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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 N/C.
Correct Answer: A — 0 N/C
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Q. If two identical charges are placed at a distance of 1m, what is the potential energy of the system?
A.
0.5 J
B.
1 J
C.
2 J
D.
4 J
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Solution
Potential energy U = k * q1 * q2 / r = (9 × 10^9 N m²/C²) * (q²) / 1m. For q = 1μC, U = 1 J.
Correct Answer: B — 1 J
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Q. If two point charges are brought closer together, what happens to the electric potential energy of the system?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Becomes zero
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Solution
The electric potential energy increases as the charges are brought closer together if they are of the same sign.
Correct Answer: A — Increases
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Q. In a capacitor, if the plate area is increased while keeping the separation constant, what happens to the capacitance?
A.
It increases
B.
It decreases
C.
It remains the same
D.
It becomes zero
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Solution
Capacitance is directly proportional to the plate area A. Increasing A increases capacitance.
Correct Answer: A — It increases
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Q. In a capacitor, what does the dielectric constant represent?
A.
The ability to store charge
B.
The ability to resist electric field
C.
The ability to increase capacitance
D.
The ability to conduct electricity
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Solution
The dielectric constant represents the ability of a material to increase the capacitance of a capacitor compared to a vacuum.
Correct Answer: C — The ability to increase capacitance
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Q. In a capacitor, what is the relationship between charge (Q), capacitance (C), and voltage (V)?
A.
Q = C + V
B.
Q = C * V
C.
Q = V / C
D.
Q = C - V
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Solution
The relationship is given by Q = C * V, where Q is the charge, C is the capacitance, and V is the voltage.
Correct Answer: B — Q = C * V
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Q. In a circuit, a capacitor is charged and then discharged through a resistor. What is the time constant of the circuit?
A.
RC
B.
C/R
C.
R/C
D.
1/RC
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Solution
The time constant (τ) of an RC circuit is given by τ = R * C.
Correct Answer: A — RC
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Q. In a circuit, if a capacitor is fully charged, what is the voltage across it?
A.
Zero
B.
Equal to the source voltage
C.
Half of the source voltage
D.
Double the source voltage
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Solution
When a capacitor is fully charged, the voltage across it is equal to the voltage of the source it was connected to.
Correct Answer: B — Equal to the source voltage
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Q. In a circuit, two capacitors of capacitance 2μF and 3μF are connected in parallel. What is the total capacitance?
A.
5μF
B.
6μF
C.
1.2μF
D.
0.6μF
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Solution
The total capacitance C_total in parallel is the sum of individual capacitances: C_total = C1 + C2 = 2μF + 3μF = 5μF.
Correct Answer: A — 5μF
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Q. In a circuit, two capacitors of capacitance 3μF and 6μF are connected in parallel. What is the total capacitance?
A.
9μF
B.
2μF
C.
18μF
D.
1μF
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Solution
In parallel, the total capacitance is the sum: C_total = C1 + C2 = 3μF + 6μF = 9μF.
Correct Answer: A — 9μF
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Q. In a circuit, two capacitors of capacitance 4μF and 6μF are connected in parallel. What is the total capacitance?
A.
10μF
B.
24μF
C.
2.4μF
D.
0.4μF
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Solution
For capacitors in parallel, the total capacitance is the sum of the individual capacitances: C_total = C1 + C2 = 4μF + 6μF = 10μF.
Correct Answer: A — 10μF
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Q. In a parallel combination of capacitors, how is the total capacitance calculated?
A.
C_eq = C1 + C2 + C3
B.
1/C_eq = 1/C1 + 1/C2 + 1/C3
C.
C_eq = 1/(C1 + C2 + C3)
D.
C_eq = C1 * C2 * C3
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Solution
In a parallel combination, the total capacitance is simply the sum of the individual capacitances: C_eq = C1 + C2 + C3.
Correct Answer: A — C_eq = C1 + C2 + C3
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Q. In a parallel plate capacitor, if the area of the plates is doubled while keeping the separation constant, what happens to the capacitance?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
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Solution
Capacitance is directly proportional to the area of the plates. Doubling the area will double the capacitance.
Correct Answer: A — It doubles
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Q. In a parallel plate capacitor, if the distance between the plates is doubled while keeping the charge constant, what happens to the electric potential?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
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Solution
The electric potential V is directly proportional to the distance d between the plates, so if d is doubled, V also doubles.
Correct Answer: A — It doubles
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Q. In a parallel plate capacitor, if the distance between the plates is halved, what happens to the capacitance?
A.
It halves
B.
It doubles
C.
It remains the same
D.
It quadruples
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Solution
Capacitance C = ε₀ * A / d. If d is halved, C doubles.
Correct Answer: B — It doubles
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Q. In a region of space where the electric field is uniform, what is the electric flux through a surface area A oriented perpendicular to the field?
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Solution
The electric flux Φ through a surface area A in a uniform electric field E is given by Φ = EA when the surface is perpendicular to the field.
Correct Answer: A — EA
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Q. In a region where the electric field is uniform, how does the electric flux through a surface depend on the angle between the field and the normal to the surface?
A.
It is maximum when the angle is 0°
B.
It is maximum when the angle is 90°
C.
It is independent of the angle
D.
It is zero when the angle is 0°
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Solution
The electric flux is maximum when the angle between the electric field and the normal to the surface is 0°, as Φ = E·A·cos(θ).
Correct Answer: A — It is maximum when the angle is 0°
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Q. In a region where the electric field is uniform, what is the shape of the Gaussian surface that would yield the simplest calculation of electric flux?
A.
Sphere
B.
Cube
C.
Cylinder
D.
Plane
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Solution
A plane is the simplest Gaussian surface for uniform electric fields, as it allows for straightforward calculation of flux.
Correct Answer: D — Plane
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Q. In a uniform electric field, how does the electric potential change with distance?
A.
Linearly
B.
Quadratically
C.
Exponentially
D.
Remains constant
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Solution
In a uniform electric field, the electric potential changes linearly with distance.
Correct Answer: A — Linearly
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Q. In a uniform electric field, the equipotential surfaces are always:
A.
Perpendicular to the field lines
B.
Parallel to the field lines
C.
Curved
D.
None of the above
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Solution
Equipotential surfaces are always perpendicular to electric field lines.
Correct Answer: A — Perpendicular to the field lines
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Q. In a uniform electric field, the equipotential surfaces are:
A.
Curved
B.
Straight lines
C.
Concentric circles
D.
Parallel planes
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Solution
In a uniform electric field, the equipotential surfaces are parallel planes.
Correct Answer: D — Parallel planes
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Q. In a uniform electric field, the potential difference between two points is directly proportional to what?
A.
Distance between the points
B.
Magnitude of the electric field
C.
Both A and B
D.
None of the above
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Solution
The potential difference is directly proportional to both the distance between the points and the magnitude of the electric field.
Correct Answer: C — Both A and B
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Q. In a uniform electric field, the potential difference between two points is given by which formula?
A.
V = Ed
B.
V = E/d
C.
V = d/E
D.
V = E × d
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Solution
In a uniform electric field, the potential difference (V) between two points is given by V = Ed, where d is the distance between the points.
Correct Answer: A — V = Ed
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Q. In a uniform electric field, the potential difference between two points is given by which of the following?
A.
E × d
B.
E/d
C.
d/E
D.
E + d
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Solution
In a uniform electric field, the potential difference (V) between two points separated by a distance (d) is given by V = E × d, where E is the electric field strength.
Correct Answer: A — E × d
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