Question

A 160 pF capacitor and a 640 pF capacitor are both charged to 1.8 kV. They are then disconnected from the voltage source and are connected together, positive plate to positive plate and negative plate to negative plate.
(a) Find the resulting potential difference across each capacitor.

(b) Find the energy lost when the connections are made.

Answer 1

Answer:

a). The resulting potential difference across each capacitor is 1.8 kV.

(b). The energy lost  is zero.

Explanation:

Given that,

First capacitor = 160 pF

Second capacitor = 640 pF

Voltage = 1.8 kV

(a). We need to calculate the initial charge

Using formula of charge

$q=CV$

Put the value into the formula

$q=160\times10^{-12}\times1.8\times10^{3}$

$q=2.88\times10^{-7}\ C$

$q=288\ nC$

We need to calculate the final charge

Using formula of charge

$q=CV$

$q=640\times10^{-12}\times1.8\times10^{3}$

$q=1152\ nC$

The total charge is

$Q=288+1152$

$Q=1440\ nC$

We need to calculate the resulting potential difference across each capacitor

Using formula of potential difference

$V=\dfrac{Q}{C_{1}+C_{2}}}$

Put the value into the formula

$V=\dfrac{1440\times10^{-9}}{(160+640)\times10^{-12}}$

$V=1.8\ kV$

(b). We need to calculate the energy lost when the connections are made

Using formula of energy

$E=\dfrac{1}{2}CV^2-\dfrac{1}{2}CV^2$

Put the value into the formula

$E=\dfrac{1}{2}\times(160+640)\times10^{-12}\times1800^2-\dfrac{1}{2}\times(160+640)\times10^{-12}\times1800^2$

$E=0$

Hence, (a). The resulting potential difference across each capacitor is 1.8 kV.

(b). The energy lost  is zero.