A parallel plate capacitor with cross-sectional area A and separation d has air between the plates. An insulat

Question

A parallel plate capacitor with cross-sectional area A and separation d has air between the plates. An insulating slab of the same area but the thickness of d/2 is inserted between the plates as shown in the figure, having a dielectric constant, K=4. The ratio of the new capacitance to its original capacitance will be:

Accepted Answer

1. **Original Capacitance:** With air between the plates, the capacitance is $C_0 = \frac{\varepsilon_0 A}{d}$.
2. **New Capacitance:** When a dielectric slab of thickness $t$ and dielectric constant $K$ is inserted, the new capacitance is given by: 
   $C' = \frac{\varepsilon_0 A}{(d-t) + \frac{t}{K}}$
3. **Substitute Values:** Here, $t = \frac{d}{2}$ and $K = 4$. 
   $C' = \frac{\varepsilon_0 A}{(d - \frac{d}{2}) + \frac{d/2}{4}} = \frac{\varepsilon_0 A}{\frac{d}{2} + \frac{d}{8}}$
4. **Simplify:** 
   The denominator is $\frac{4d + d}{8} = \frac{5d}{8}$.
   $C' = \frac{\varepsilon_0 A}{\frac{5d}{8}} = \frac{8}{5} \left( \frac{\varepsilon_0 A}{d} \right) = \frac{8}{5} C_0$
5. **Ratio:** The ratio $\frac{C'}{C_0} = \frac{8}{5}$.

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