A parallel-plate capacitor having plate area 100cm² and separation 1.0 mm holds a charge of $0 .12 μC$ when connected to a 120 V battery. Find the dielectric constant of the material filling the gap.

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

11.3

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

The capacitance of the capacitor is $\frac{0.12 μ C}{120 V} = 1.0 \times 10^{- 9} F$
If K be the dielectric constant, the capacitance is also given by $\frac{{Kε}_{0} A}{d}$ .

Thus,
$\frac{K \times 8 .85 \times 10^{- 12} \times 100 \times 10^{- 4}}{1 .0 \times 10^{- 3}} = 1 .0 \times 10^{- 9}$
K = 11.3

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!