AP EAMCET · PHYSICS · Capacitance

A $60 μF$ parallel plate capacitor whose plates are separated by $6 mm$ is charged to $250 V$ , and then the charging source is removed. When a slab of dielectric constant $5$ and thickness $3 mm$ is placed between the plates, find the change in the potential difference across the capacitor?

A $250 V$
B $100 V$
C $150 V$
D $75 V$

Verified Solution

Answer & Solution

Correct Answer

(B) $100 V$

Step-by-step Solution

Detailed explanation

Capacitance of capacitor after inserting a slab of dielectric constant 5 and of thickness t=3 mm is, C'=C11-td+tkd ⇒C'=12 μF11-3 mm6 mm+3 mm5×6 mm=12 μF×2012⇒C'=20 μF Now, after removing the…

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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A 60 μF parallel plate capacitor whose plates are separated by 6 mm is charged to 250 V, and then the charging source is removed. When a slab of dielectric constant 5 and thickness 3 mm is placed between the plates, find the change in the potential difference across the capacitor?

Step-by-step Solution

A $60 μF$ parallel plate capacitor whose plates are separated by $6 mm$ is charged to $250 V$ , and then the charging source is removed. When a slab of dielectric constant $5$ and thickness $3 mm$ is placed between the plates, find the change in the potential difference across the capacitor?