A uniformly charged solid sphere of radius R has potential V 0 (measured with respect to ∞ ) on its…

A uniformly charged solid sphere of radius R has potential V0 (measured with respect to ∞ ) on its surface. For this sphere the equipotential surfaces with potential 3V02,5V04,3V04 and V04 have radius R1 , R2, R3 and R4 respectively. Then

Note : This question had two option correct at the time of examination. Proper corrections are made in the question to avoid it.

  1. 2R>R4
  2. R1=0 and R2>R4-R3
  3. R1≠0 and R2-R1> (R4-R3) 
  4. R1=0 and R2<R4-R3

Solution


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 < 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
 

Asked in: JEE Main 2015 (04 Apr)

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