Question

The capacity of a parallel plate capacitor with no dielectric substance but with a separation of 0.4 cm is $2\mu F$. The separation is reduced to half and it is filled with a dielectric substance of value 2.8. The final capacity of the capacitor is

• A. $11.2\mu F$
• B. $15.6\mu F$
• C. $19.2\mu F$
• D. $22.4\mu F$

Answer 1

The correct option is A $11.2\mu F$

Initial capacitance $C=2uF=\frac{{\epsilon }_{0}A}{d}$
New capacitance $C^{\prime }=\frac{K{\epsilon }_{0}A}{d^{\prime }}=\frac{2.8\times {\epsilon }_{0}A}{\left(d/2\right)}=5.6\frac{{\epsilon }_{0}A}{d}$
$\Rightarrow C^{\prime }=5.6\times 2=11.2\mu F$