Question

A slab of copper of thickness b is inserted in between the plates of parallel plate capacitor as shown in the figure. The separation between the plates is d. If $b = \frac { d } { 2 }$ then the ratio of capacities of the capacitor after and before inserting the slab will be

• A. $\sqrt { 2 } : 1$
• B. 2 : 1
• C. 1 : 1
• D. $1 : \sqrt { 2 }$

Answer 1

Answer: (B)

2 : 1

Answer 2

Capacitance before inserting the slab $C ^ { \prime } = \frac { \varepsilon _ { 0 } A } { d - t }$.

Where $C ^ { \prime } = \frac { 2 \varepsilon _ { 0 } A } { d }$ hence,