Solveeit Logo

Question

Physics Question on electrostatic potential and capacitance

Figure shows three spherical and equipotential surfaces A, B and C round a point charge The potential difference VAVB=VBVC.Ift1andt2V_A - V_B = V_B - V_C. \, If \, t_1 \, and \, t_2 be the distances between them, then

A

t1=t1 t_1 = t_1

B

t1>t2 t_1 > t_2

C

t1<t2 t_1 < t_2

D

t1t2 t_1 \le t_2

Answer

t1<t2 t_1 < t_2

Explanation

Solution

PotentiaI difference between two equipotential surfaces A
and B.
VAVB=kq(1rA1rB)=kq(rBrArArB)=kqt1rArBV_A - V_B = kq \bigg( \frac{ 1}{ r_A } - \frac{ 1}{ r_B} \bigg) = kq \bigg( \frac{ r_B - r_A }{ r_A \, r_B} \bigg) = \frac{ kq t_1 }{ r_A \, r_B}
or t1=(VAVB)rArBkq t_1 = \frac{ ( V_A - V_B) \, r_A \, r_B}{ kq}
or t1rArB t_1 \propto r_A \, r_B
Simiiarly, t1rBrC t_1 \propto r_B \, r_C
Since, rA<rB<rC,thereforerArB<rBrC r_A < r_B < r_C, \, therefore \, r_A \, r_B < r_B \, r_C
t1<t2\therefore t_1 < t_2