A uniformly charged solid sphere of radius R has potential V 0 (measured with respect to ∞ ) on its surface. For this sphere the equipotential surfaces with potential 3 V 0 2 , 5 V 0 4 , 3 V 0 4 and V 0 4 have radius R 1 , R 2 , R 3 and R 4 respectively. Then Note : This question had two option correct at the time of examination. Proper corrections are made in the question to avoid it.

Question

A uniformly charged solid sphere of radius R has potential V 0 (measured with respect to ∞ ) on its surface. For this sphere the equipotential surfaces with potential 3 V 0 2 , 5 V 0 4 , 3 V 0 4 and V 0 4 have radius R 1 , R 2 , R 3 and R 4 respectively. Then Note : This question had two option correct at the time of examination. Proper corrections are made in the question to avoid it., 2 R > R 4, R 1 = 0 and R 2 > R 4 - R 3, R 1 ≠ 0 and R 2 - R 1 > ( R 4 - R 3 ), R 1 = 0 and R 2 R 4 - R 3

MARKS App · Accepted Answer

Potential for uniformly charged solid sphere v = 1 4 π ε 0 Q r outside i.e r > R v = 1 4 π ε 0 Q R on the surface v = 1 4 π ε 0 Q R 3 2 - 1 2 r 2 R 2 inside i.e. r Clearly 3 v 0 2 , 5 v 0 4 are inside potentials ∵ > v 0 3 v 0 4 , v 0 4 are outside potentials ∵ v 0 To get R 1 : 3 v 0 2 = 1 4 π ε 0 Q R 3 2 - 1 2 R 1 2 R 2 where, v 0 = 1 4 π ε 0 Q R 3 2 × 4 π ε 0 Q R = 1 4 π ε 0 Q R 3 2 - 1 2 R 1 2 R 2 3 2 = 3 2 - 1 2 R 1 2 R 2 ⇒ R 1 = 0 To get R 2 : 5 4 v 0 = 1 4 π ε 0 Q R 3 2 - 1 2 R 2 2 R 2 5 4 1 4 π ε 0 Q R = 1 4 π ε 0 Q R 3 2 - 1 2 R 2 2 R 2 5 4 = 3 2 - 1 2 R 2 2 R 2 1 2 R 2 2 R 2 = 1 4 ⇒ R 2 = R 2 To get R 3 : 3 v 0 4 = 1 4 π ε 0 Q R 3 3 4 1 4 π ε 0 Q R = 1 4 π ε 0 Q R 3 3 4 R = 1 R 3 R 3 = 4 3 R To get R 4 : v 0 4 = 1 4 π ε 0 Q R 4 1 4 × 1 4 π ε 0 Q R = 1 4 π ε 0 Q R 4 R 4 = 4 R R 4 - R 3 = 4 R - 4 R 3 = 8 R 3 > R 2 R 1 = 0 and R 2 R 4 - R 3

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