Question

4..n identical small spherical drops, each of radius r are charged to same potential V .They are combined to form a bigger drop. The potential of the big drop will be ..........​

Answer 1

Answer:

n^5/3E•

Explanation:

POTTENTIAL V=q/4PIE E•r r=radius of droplet

n×4/3pie r^3=4/3R^2 R=RADIUS OF THE BIG DROPLET

R=n^1/3r

Pottential V=nq/4pieE•aR=nq/4PIE E•n^1/3r=n^2/3V

E•=1/2q^2/4PIE E•r

E•=1/2(nq)^2/4PIE E•RA

=1/2q^2/4 PIE E•r×n^1/3

=n^5/3E•

Where E•=EPSILONNOT

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