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 potentials 3V02,5V04,3V04 and V04 have radius R1,R2,R3 and R4 respectively. Then :
R1=0 and R2>(R4−R3)
R1≠0 and (R2−R1)>(R4−R3)
2R<R4
None of the above
A
R1=0 and R2>(R4−R3)
B
2R<R4
C
None of the above
D
R1≠0 and (R2−R1)>(R4−R3)
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Solution
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V(R)=kQR=V0
We use results derived for potential.
V(r)=kQr for r>R
V(r)=kQ2R3(3R2−r2)
V(r=0)=3kQ2R=3V02
Hence, V(R1) matches with V at center.
∴R1=0
Thus, we eliminate option(B).
At R2
V(R2)=5V04=kQ2R3(3R2−R22)
⇒5V04=12(kQR)(3−R22R2)=12V0(3−R22R2)
⇒52=3−R22R2
⇒12=R22R2
⇒R2=R√2
For R3
V(R3)=3V04=kQR3
⇒3V04=V0RR3
⇒R3=4R3
For R4
V(R4)=V04=kQR4
⇒R4=4R
Thus, R4−R3=4R−4R3=8R3
R2=R√2<8R3=R4−R3
Also, 2R<4R=R4
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