Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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:

1:2

2:3

3:2

2:1

Step-by-Step Solution

Formula for Potential: For a parallel plate capacitor with charge $Q$ , plate area $A$ , and separation $d$ , the potential difference is $V_0 = \frac{Qd}{\varepsilon_0 A}$ .
With Dielectric: When a dielectric slab of thickness $t$ and dielectric constant $K$ is inserted, the new potential difference $V'$ is given by $V' = \frac{Q}{\varepsilon_0 A} \left[ (d-t) + \frac{t}{K} \right]$ .
Given Values: $K = 3$ , $t = \frac{3}{4}d$ .
Substitute: $V' = \frac{Q}{\varepsilon_0 A} \left[ (d - \frac{3}{4}d) + \frac{3d/4}{3} \right]$ $V' = \frac{Q}{\varepsilon_0 A} \left[ \frac{d}{4} + \frac{d}{4} \right] = \frac{Q}{\varepsilon_0 A} \left[ \frac{d}{2} \right] = \frac{V_0}{2}$ .
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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