Question

Following figure shows the initial charge on the capacitors . After the switch S is closed , find-
Q1. charge on C1
Q2. charge on C2
Q3. work done by battery

Answer 1

Given that,

First capacitor = 1 μF

Second capacitor = 1 μF

Voltage = 10 V

(I). We need to calculate the charge on C₁

Using formula of charge

$Q_{1}=C_{1}V$

Put the value into the formula

$Q_{1}=1\times10$

$Q_{1}=10\ \mu C$

(II). We need to calculate the charge on C₂

Using formula of charge

$Q_{2}=C_{2}V$

Put the value into the formula

$Q_{2}=1\times10$

$Q_{2}=10\ \mu C$

We need to calculate the total capacitor

Using formula of series

$\dfrac{1}{C}=\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}}$

Put the value into the formula

$\dfrac{1}{C}=\dfrac{1}{1}+\dfrac{1}{1}$

$C=\dfrac{1}{2}\ \mu F$

We need to calculate the total charge

Using formula for charge

$Q=Q_{1}+(-Q_{2})$

Put the value into the formula

$Q=10-10$

$Q=0$

(III). We need to calculate the work done by battery

Using formula of work done

$W=\dfrac{1}{2}\dfrac{Q^2}{C}$

Put the value in to the formula

$W=\dfrac{1}{2}\times\dfrac{2\times0}{1}$

$W=0\ J$

Hence, (I). The charge on C₁ is 10 μC.

(C) is correct option.

(II). The charge on C₂ is 10 μC

(C) is correct option.

(III). The work done by battery is zero.

(D) is correct option.

Answer 2

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Given that,

First capacitor = 1 μFSecond capacitor = 1 μF

Voltage = 10 V(I). We need to calculate the charge on C₁

Using formula of chargeQ_{1}=C_{1}VQ 1

Put the value into the formula

Q_{1}=1\times10Q

(II). We need to calculate the charge on C₂Using formula of chargePut the value into the formulaWe need to calculate the total capacitor

Using formula of

Put the value into the We need to calculate the total chargeUsing formula for chargeQ=Q_{1}+(-Q_{2})Q=Q Put the value into the formulaQ=10-10Q=10−10Q=0Q=0

W=0\ JW=0 JHence, (I). The charge on C₁ is 10 μC.

(C) is correct option.(II). The charge on C₂ is 10 μC

(C) is correct option.

(III). The work done by battery is zero.

(D) is correct option

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