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Try this! - Electrostatics - JEE Main-9

In the given circuit, the charge on 4\mu F the capacitor will be:

 

  • Option 1)

    5.4 \mu C

     

  • Option 2)

    9.6 \mu C

  • Option 3)

    13.4 \mu C

  • Option 4)

    24 \mu C

 
Answers (1)
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Capacitance of Conductor -

Q\propto V

Q=CV

- wherein

C - Capacity or capacitance of conductor 

V - Potential.

 

 

\frac{1}{C_{er}}=\frac{1}{4}+\frac{1}{5+1}=\frac{1}{4}+\frac{1}{6}=\frac{10}{24}=\frac{5}{12}

C_{er}=\frac{12}{5}

Q=CV 

\frac{12}{5}\times 10=24\mu C


Option 1)

5.4 \mu C

 

Option 2)

9.6 \mu C

Option 3)

13.4 \mu C

Option 4)

24 \mu C

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