A parallel plate capacitor is connected with a 1438 volt bat

A parallel plate capacitor is connected with a 1.438 volt battery and each plate contains 2,450 micro Coulomb charge. How much energy is stored in the capacitor? Answer: A certain capacitor stores 37J of energy when it holds 2.650 uC of charge. What is the capacitance in nF? Answer:

Solution

Potential difference V = 1438 volt

Charge q = 2450 x10 ^-6 Coulomb

Energy stored in the capacitor U = (1/2)qV

                                                = 0.5 x 1438 x(2450 x10 ^-6 )

                                               = 1.76155 J

(b). Energy stored U \' = 37 J

Charge q \' = 2650 x10 ^-6 Coulomb

Capacitance C = ?

We know U \' =q \' ² /2C

From this capacitance C = q \' ²/2U \'

                                    = (2650x10 ^-6) ²/(2 x37)

                                    = 94.89 x10 ^-9 F

                                    = 94.89 nF