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  • A parallel plate capacitor having cross sectional area A and separation d has air in between the plates . Now an insulating slab of same area but thickness d/2 is inserte

A parallel plate capacitor having cross sectional area A and separation d has air in between the plates . Now an insulating slab of same area but thickness d/2 is inserted between the plates s shown in figure having dielectric constant K(=4). The ratio of new capacitance to its original capacitance will be.

Option: 1

8:5


Option: 2

6:5


Option: 3

4:1


Option: 4

2:1


Answers (1)

best_answer

C=\frac{A\varepsilon _0}{\frac{t_1}{K_1} +\frac{t_2}{K_2}}

C_{new}=\frac{A\varepsilon _0}{\frac{\frac{d}{2}}{1}+\frac{\frac{d}{2}}{4}}= \frac{8A\varepsilon _0}{5d}

C_{original}=\frac{A\varepsilon _0}{d}

\frac{C_{new}}{C_{original}}= \frac{8}{5}

 

Posted by

Pankaj Sanodiya

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