Question

The distance between at the plates of a charged plate capacitor disconnected from the battery is 5 cm and the intensity of the field in it is E = 300 V/cm. An uncharged metal bar 1 cm thick is introduced into the capacitor parallel between its plates. If the battery provided 6 uC charge, find final capacitance.

• A. 4 $\mu f$
  
• B. 5 $\mu f$
• C. 6 $\mu f$
• D. 8 $\mu f$

Answer 1

The correct option is A 4 $\mu f$

Before inserting plate:-

Charge gained =$6\mu c$ and electric field $=300/cm$

the voltage developed across the plates,

$V=E.d=300\times 5=1500V$

Thus capacitance $c=\frac{Q}{V}=\frac{6\times {10}^{-6}}{1500}=4\times {10}^{-9}=4nF$

now the mental plate in not dieretric i.e when inserted it would not generate the reverse field and thus the capacitance wouldn't be affected

final capacitance $4\times {10}^{-9}=4nF$