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# Solve it, - Electrostatics - JEE Main-6

Condenser A has a capacity of 15 $\mu$F when it is filled with a medium of dielectric constant 15. Another condenser B has a capacity 1 $\mu$F with air between the plates. Both are charged separately by a battery of 100V. after charging, both are connected in parallel without the battery and the dielectric material being removed. The common potential now is

• Option 1)

400V

• Option 2)

800V

• Option 3)

1200V

• Option 4)

1600V

102 Views
D

Thank you, sir.

As we have learned

Parallel Grouping -

$\dpi{100} C_{eq}=C_{1}+C_{2}+\cdots$

- wherein

Charge on capacitor A is given by  $Q_1= 15*10^{-6}*100= 15*10^{-4}C$
Charge on capacitor B is given by  $Q_2= 1*10^{-6}*100= 10^{-4}C$
Capacity of capacitor A after removing dielectric  $= \frac{15*10^{-6}}{15}= 1 \mu F$
Now when both capacitors are connected in parallel their equivalent capacitance will be Ceq  = 1+1 = $2\mu F$
So common potential  $= \frac{(15*10^{-4}+1*10^{-4})}{2*10^{-6}}= 800V$

Option 1)

400V

Option 2)

800V

Option 3)

1200V

Option 4)

1600V

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