A slab of material of dielectric constant 3 has the same area as the plate of a parallel plate capacitor but has a thickness \(\left(\frac{3}{4}\right) d\), where d is the separation of the plates. What is the electrical potential difference between the plates, when the slab is inserted between the plates? Initial electrical potential difference \(V _{ 0 }\) .

A \(\frac{V_0}{6}\)
B \(\frac{V_0}{4}\)
C \(\frac{V_0}{2}\)
D \(\frac{V_0}{3}\)

Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{V_0}{2}\)

Step-by-step Solution

Detailed explanation

(C)
In absence of dielectric, electric field between two plates,
Suppose, \(E _0=\frac{ V _0}{d}\)
When dielectric is inserted, electric field inside dielectric \(E=\frac{E_0}{K}\)

So potential difference,
\(\begin{aligned} V & = E _0\left(\frac{1}{4} d\right)+\frac{ E _0}{K}\left(\frac{3}{4} d\right) \\ \therefore V & = E _0 d\left[\frac{1}{4}+\frac{3}{4 K}\right] \\ \therefore V & = V _0\left[\frac{K+3}{4 K}\right]\left(\because E _0 d= V _0\right) \\ \therefore V & = V _0\left[\frac{3+3}{4 \times 3}\right] \\ \therefore V & = V _0\left(\frac{6}{12}\right) \\ \therefore V & =\frac{ V _0}{2}\end{aligned}\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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