Question

The parallel plates in a capacitor, with a plate area of 8.50 cm2 and an air-filled separation of 3.00 mm, are charged by a 6.00 V battery. They are then disconnected from the battery and pulled apart (without discharge) to a separation of 8.00 mm. Neglect fringing. (a) Find the potential difference between the plates. (b) Find the initial stored energy.

Answer 1

Answer:

(A) Potential difference between plates will be 6 volt

(b) Initial energy stored will be $45.126\times 10^{-7}J$

Explanation:

It is given area of parallel late capacitor $A=8.50cm^2=8.5\times 10^{-4}m^2$

Initial voltage $V_0=6volt$

Distance between plates d = 3 mm = 0.003 m

Capacitance $C=\frac{\epsilon _0A}{d}=\frac{8.85\times 10^{-7}\times 8.5\times 10^{-4}}{0.003}=2.507\times 10^{-7}F$

So charge on the capacitor $q_0=CV_0=2.507\times 10^{-7}\times 6=15.042\times 10^{-7}C$

(a) This charge will be constant after disconnection.

After disconnection voltage will be

$V_1=\frac{q_0}{C}=\frac{15.042\times 10^{-7}}{2.507\times 10^{-7}}=6volt$

(b) Initial energy stored in the capacitor

$U=\frac{1}{2}CV^2=\frac{1}{2}\times 2.507\times 10^{-7}\times 6^2=45.126\times 10^{-7}J$