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In the given circuit, charge Q2 on the 2\muF capacitor changes as C is varied from 1\muF to 3\muF. Q2 as a function of ‘C’ is given properly by (figures are drawn schematically and are not to scale)

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)


Answers (2)

As we learnt in 

Direction of Electric field -

Due to Positive Charge electric field is always away from the charge.




Direction of Electric field -


Due to Negative charge electric field is always towards the charge.



 \frac{1}{C_{eq}}= \frac{1}{(1+2)} + \frac{1}{C}= \frac{1}{C}+\frac{1}{3}

C_{eq}= \frac{3C}{C+3}

Total charge in the circuit 

Q=C_{eq}E= \frac{3CE}{C+3}

Q_{2}= \frac{2}{3}Q= \frac{2}{3}\times \frac{3CE}{C+3}= \frac{2CE}{C+3}\ or\ Q_{2}= \frac{2E}{1+\frac{3}{C}}

and \frac{\partial Q_{2}}{\partial c}=\frac{6E}{(C+3)^{2}}

As C increases,  Q_{2} increases and slope of Q_{2}  -C curve decreases. Hence, Graph (d) represents the correct variation. 

Option 1)

Incorrect option

Option 2)

Correct option

Option 3)

Incorrect option

Option 4)

Incorrect option

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