Q33. A parallel plate capacitor whose capacitance C is 13.0 pF is charged by a battery to a potentia

Q: What is the correct answer for CUET UG Physics Q33?

The correct answer is A. The potential energy (U) stored in a capacitor is given by the formula U = (1/2)CV², where C is the capacitance and V is the potential difference across the capacitor. Given: Capacitance C = 13.0 pF = 13.0 x 10⁻¹² F. Potential difference V = 13.0 V. Substitute the given values into the formula: U = (1/2) * (13.0 x 10⁻¹² F) * (13.0 V)² U = (1/2) * 13.0 * 169 * 10⁻¹² J U = 0.5 * 2197 * 10⁻¹² J

Q: Q33. A parallel plate capacitor whose capacitance C is 13.0 pF is charged by a battery to a potential difference V = 13.0 V between its plates. What is the potential energy of the capacitor? (a)…

The correct answer is A. The potential energy (U) stored in a capacitor is given by the formula U = (1/2)CV², where C is the capacitance and V is the potential difference across the capacitor. Given: Capacitance C = 13.0 pF = 13.0 x 10⁻¹² F. Potential difference V = 13.0 V. Substitute the given values into the formula: U = (1/2) * (13.0 x 10⁻¹² F) * (13.0 V)² U = (1/2) * 13.0 * 169 * 10⁻¹² J U = 0.5 * 2197 * 10⁻¹² J

Q: What are the options for this CUET UG Physics question?

A. 1.098 nJ · B. 2197 pF · C. 1.98 pJ · D. 2.197 nJ

Question

Q33. A parallel plate capacitor whose capacitance C is 13.0 pF is charged by a battery to a potential difference V = 13.0 V between its plates. What is the potential energy of the capacitor?
(a) 1.098 nJ
(b) 2197 pF
(c) 1.98 pJ
(d) 2.197 nJ

FirstInTest · Accepted Answer

The potential energy (U) stored in a capacitor is given by the formula U = (1/2)CV², where C is the capacitance and V is the potential difference across the capacitor. 
Given: Capacitance C = 13.0 pF = 13.0 x 10⁻¹² F. Potential difference V = 13.0 V. 
Substitute the given values into the formula: 
U = (1/2) * (13.0 x 10⁻¹² F) * (13.0 V)² 
U = (1/2) * 13.0 * 169 * 10⁻¹² J 
U = 0.5 * 2197 * 10⁻¹² J 
U = 1098.5 * 10⁻¹² J. 
To express this in more convenient units: 10⁻¹² J = 1 pJ (picojoule) and 10⁻⁹ J = 1 nJ (nanojoule). So, U = 1098.5 pJ = 1.0985 nJ. 
Rounding to three significant figures, U ≈ 1.098 nJ. Option (a) matches this value. Note that option (b) gives units of pF, which is a unit of capacitance, not energy.

Answer

1.098 nJ

Answer

2197 pF

Answer

1.98 pJ

Answer

2.197 nJ

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