Q. A capacitor is charged to a potential difference of 12V. If it is disconnected from the battery and the plates are moved apart, what happens to the potential difference? (2021)
A.Increases
B.Decreases
C.Remains the same
D.Becomes zero
Solution
When the plates of a disconnected capacitor are moved apart, the capacitance decreases, leading to an increase in potential difference.
Q. A capacitor is charged to a potential of 12V and then disconnected from the battery. If the distance between the plates is doubled, what is the new potential difference? (2022)
A.6V
B.12V
C.24V
D.0V
Solution
When the distance is doubled, the potential difference across the capacitor also doubles, resulting in 24V.
Q. A capacitor is charged to a potential of 12V and then disconnected from the battery. If the plate area is doubled, what will be the new potential difference? (2022)
A.6V
B.12V
C.24V
D.It cannot be determined
Solution
Once disconnected, the charge remains constant. Doubling the area increases capacitance but does not change the potential difference since the charge is fixed.
Q. A capacitor is charged to a voltage of 12V and then disconnected from the battery. If the distance between the plates is doubled, what happens to the voltage across the capacitor? (2023)
A.It remains the same
B.It doubles
C.It halves
D.It becomes zero
Solution
When the distance between the plates is doubled, the capacitance decreases, which causes the voltage to double since Q remains constant.
Q. A capacitor of capacitance C is charged to a voltage V. If the charge is then removed, what is the potential difference across the capacitor? (2023)
A.0
B.V
C.C
D.CV
Solution
If the charge is removed, the potential difference across the capacitor becomes 0 volts.
Q. If a capacitor is charged to a voltage of 5 V and then disconnected from the battery, what happens to the charge on the capacitor if the voltage is increased to 10 V? (2023)
A.Charge increases
B.Charge decreases
C.Charge remains the same
D.Charge becomes zero
Solution
When the voltage is increased to 10 V, the charge on the capacitor increases because charge (Q) is directly proportional to voltage (V) for a given capacitance.
Q. If the distance between two point charges is tripled, how does the electrostatic force between them change? (2020)
A.It triples
B.It halves
C.It becomes one-ninth
D.It remains the same
Solution
According to Coulomb's law, the force F is inversely proportional to the square of the distance (F ∝ 1/r^2). If r is tripled, F becomes 1/9 of its original value.
Q. In a capacitor, if the plate area is increased while keeping the distance constant, what happens to the capacitance? (2023)
A.Increases
B.Decreases
C.Remains the same
D.Becomes zero
Solution
Increasing the plate area while keeping the distance constant increases the capacitance, as capacitance is directly proportional to the area of the plates.
Q. The capacitance of a parallel plate capacitor is directly proportional to which of the following? (2019)
A.Distance between plates
B.Area of plates
C.Voltage across plates
D.Dielectric constant
Solution
Capacitance (C) is directly proportional to the area of the plates (A) and the dielectric constant (ε) and inversely proportional to the distance (d) between the plates.
Q. The capacitance of a parallel plate capacitor is given by which formula? (2020)
A.C = ε₀A/d
B.C = Q/V
C.C = 1/(4πε₀r)
D.C = V/I
Solution
The capacitance (C) of a parallel plate capacitor is given by the formula C = ε₀A/d, where A is the area of the plates and d is the separation between them.
Q. The electric field between the plates of a parallel plate capacitor is given by which expression? (2020)
A.E = V/d
B.E = d/V
C.E = CV
D.E = Q/A
Solution
The electric field (E) between the plates of a parallel plate capacitor is given by the expression E = V/d, where V is the voltage and d is the distance between the plates.
Q. The electric field between the plates of a parallel plate capacitor is given by which formula? (2020)
A.E = V/d
B.E = Q/A
C.E = CV
D.E = 1/(4πε₀r²)
Solution
The electric field (E) between the plates of a parallel plate capacitor is given by the formula E = V/d, where V is the voltage and d is the separation between the plates.
