Question

A 2uF capacitor is charged to 200 volt and then the battery is
disconnected When it is connected in parallel to another uncharged
capacitor the potential difference between the plates of both is 40
volt The capacitance of the other capacitor is​

Answer 1

The capacitance of the other capacitor is​ $8\ \mu F$

Explanation:

Given that,

Capacitor = 2μF

Voltage = 200 volt

We need to calculate the charge

Using formula of charge

$Q=CV$

Where, V = potential

C = capacitor

put the value into the formula

$Q=2\times10^{-6}\times200$

$Q=400\times10^{-6}\ C$

$Q=400\ \mu C$

We need to calculate the other capacitor

Using conservation of charge

$Q_{in}=Q_{final}$

$400\times10^{-6}=2\times10^{-6}\times40+C\times40$

$C=\dfrac{400\times10^{-6}-2\times10^{-6}\times40}{40}$

$C=0.000008\ F$

$C=8\times10^{-6}\ F$

$C=8\ \mu F$

Hence, The capacitance of the other capacitor is​ $8\ \mu F$

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Topic : capacitor

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