Question

A parallel-plate capacitor with plates of area 600 cm^2 is charged to a potential difference V and is then disconnected from the voltage source. When the plates are moved 0.7 cm farther apart, the voltage between the plates increases by 100 V.(a) What is the charge Q on the positive plate of the capacitor?_________nC(b) How much does the energy stored in the capacitor increase due to the movement of the plates?_________µJ

Answer 1

Answer:

a) Q = 0.759µC

b) E = 39.5µJ

Explanation:

a) The charge Q on the positive charge capacitor can be gotten using the formula Q = CV

C = capacitance of the capacitor (in Farads )

V = voltage (in volts) = 100V

C = ∈A/d

∈ = permittivity of free space = 8.85 × 10^-12 F/m

A = cross sectional area = 600 cm²

d= distance between the plates = 0.7cm

C = 8.85 × 10^-12 * 600/0.7

C = 7.59*10^-9Farads

Q = 7.59*10^-9 * 100

Q = 7.59*10^-7Coulombs

Q = 0.759*10^-6C

Q = 0.759µC

b) Energy stored in a capacitor is expressed as E = 1/2CV²

E = 1/2 * 7.59*10^-9 * 100²

E = 0.0000395Joules

E = 39.5*10^-6Joules

E = 39.5µJ