Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5

Q.

Statement (A): Inside a charged hollow metal sphere, E = 0, V 0. (E-electric field, V-electric potential) 

Statement (B): The work done in moving a positive charge on a equipotential surface is zero. 

Statement (C): When two like charges are brought closer, their mutual electrostatic potential energy will increase.

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

a

A, B, C are true

b

A, B are true, C is false

c

A, C are true, B is false

d

B, C are true, A is false

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

Inside a hollow metal sphere E=0 and V=KQR0

So, (A) is true,

No work is done in moving a charge from one point to other on equipotential
surface as ΔV=0 on equipotential
surface. So W = q
ΔV=0

So, (B) is true.

When two like charges are brought closer, then external agent do the work against the mutual repulsion. So potential energy of system increases. 

So, (C) is true.

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon
Statement (A): Inside a charged hollow metal sphere, E = 0, V≠ 0. (E-electric field, V-electric potential) Statement (B): The work done in moving a positive charge on a equipotential surface is zero. Statement (C): When two like charges are brought closer, their mutual electrostatic potential energy will increase.