PYQ NEET- Electrostatic Potential And Potential Energy

1. A parallel plate capacitor of capacitance C is charged to a potential V. The distance between the plates is then doubled. The ratio of the new capacitance to the old capacitance is

C/2 C/4 C/8 (D) C/16

The answer is (B). The capacitance of a parallel plate capacitor is given by C = εA/d, where ε is the permittivity of the dielectric between the plates, A is the area of the plates, and d is the distance between the plates. When the distance between the plates is doubled, the capacitance is reduced by a factor of 1/2.

2. The electric field at a point near a charged conductor is 50 N/C. If the potential at that point is 300 V, the magnitude of the charge on the conductor is

(A) 15 μC (B) 60 μC (C) 120 μC (D) 180 μC

The answer is (B). The electric field at a point near a charged conductor is given by E = σ/ε₀