Question

A parallel plate capacitor is connected to a battery of 10 V and can store a charge of 50 μC. What will be the capacitance of the parallel plate capacitor. 11. Two metal plates of dimensions 15 cm x 10 cm are kept at a distance of 2 mm in air. What will be the capacitance of the parallel plate capacitor formed by the plates when it is connected to a battery of 10 V? In the previous situation, calculate the capacitance of the parallel plate capacitor if a dielectric slab of dielectric constant 2.5 and thickness 1 mm is inserted between the plates when it is connected to a battery of 10 V.​

Answer 1

Explanation:

Q = CV

50 × 10^(-6) = C × 10

C = 50 × 10^(-7) Farad

C = Ae●/d

C = [15 × 10^(-2) × 10 × 10^(-2) × 8.85 × 10^(-12)]/[2 × 10^(-3)]

For dielectric,

C = kAe●/d 【k = dielectric constant】