Figure shows three spherical and equipotential surfaces A, B and C around a point charge q. The potential difference VA−VB=VB−VC. If t1 and t2 be the distances between them, then :
t1=t2
t1≤t2
t1<t2
t1>t2
A
t1=t2
B
t1>t2
C
t1≤t2
D
t1<t2
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Solution
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The potential around a point charge is V(r)=q4πϵ0r
Let a,b,c be the radii of A,B,C respectively.
Thus, t1=b−a and t2=c−b
VA−VB=q4πϵ0a−q4πϵ0b=qt14πϵ0ab
VB−VC=q4πϵ0b−q4πϵ0c=qt24πϵ0bc
But, VA−VB=VB−VC
⇒VA−VBVB−VC=1⇒t1ab×bct2=1⇒t1t2=ac
As a<c, we have t1<t2
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