A dielectric slab of dielectric constant 3 having the same area of cross-section as that of a parallel plate c

Question

A dielectric slab of dielectric constant 3 having the same area of cross-section as that of a parallel plate capacitor but of thickness 3/4th of the separation of the plates is inserted into the capacitor. The ratio of potential difference across the plates without dielectric to that with dielectric is:

Accepted Answer

1. **Formula for Potential:** For a parallel plate capacitor with charge $Q$, plate area $A$, and separation $d$, the potential difference is $V_0 = \frac{Qd}{\varepsilon_0 A}$.
2. **With Dielectric:** When a dielectric slab of thickness $t$ and dielectric constant $K$ is inserted, the new potential difference $V'$ is given by $V' = \frac{Q}{\varepsilon_0 A} \left[ (d-t) + \frac{t}{K} \right]$.
3. **Given Values:** $K = 3$, $t = \frac{3}{4}d$.
4. **Substitute:**
   $V' = \frac{Q}{\varepsilon_0 A} \left[ (d - \frac{3}{4}d) + \frac{3d/4}{3} \right]$
   $V' = \frac{Q}{\varepsilon_0 A} \left[ \frac{d}{4} + \frac{d}{4} \right] = \frac{Q}{\varepsilon_0 A} \left[ \frac{d}{2} \right] = \frac{V_0}{2}$.
5. **Ratio:** The ratio of potential difference without dielectric ($V_0$) to that with dielectric ($V'$) is:
   $\frac{V_0}{V'} = \frac{V_0}{V_0 / 2} = 2:1$.

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