A potential difference of 2000 V is established between the parallel plates of a capacitor with air between the plates. If the air becomes conducting when the electric field exceeds 3 x 106 N/C, the minimum separation between the plates should be

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

0 .01m

0.001m

0. 667 mm

1.34 mm

answer is C.

(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

Let the minimum separation should be x. Then E=Vd=2000x=3×106∴ x=20003×106=23×10−3=0.667mm

Watch 3-min video & get full concept clarity