Question

# A solid conducting sphere having a charge $$Q$$ is surrounded by an uncharged concentric conducting spherical shell. The potential difference between the surface of solid sphere and the shell is $$V$$. The shell is now given a charge $$-3Q$$. The new potential difference between the same surfaces will be :

A
2V
B
4V
C
V
D
2V

Solution

## The correct option is C $$V$$A solid conducting sphere having a charge Q is surrounded by an uncharged concentric spherical shell, The potential difference between the solid sphere and hallow shell is $$V$$.$$V_A-V_B=V$$$$V_A-V_B=\dfrac{Q}{4\pi \varepsilon_0 a}.\dfrac{Q}{4\pi \varepsilon_0 .b}$$$$V=\dfrac{Q}{4\pi \varepsilon_0}[\dfrac{1}{a}-\dfrac{1}{b}]$$. . . . . . .(1)Now, shell is given a charge $$-3Q$$,the,The new potential  is,$$V'_A=\dfrac{Q}{4\pi \varepsilon_0.a}-\dfrac{3Q}{4\pi \varepsilon_0.b}$$$$V'_A=\dfrac{Q}{4\pi \varepsilon_0}[\dfrac{1}{a}-\dfrac{3}{b}]$$ and,$$V'_B=\dfrac{Q}{4\pi \varepsilon_0 b}-\dfrac{3Q}{4\pi \varepsilon_0 b}$$$$V'_B=\dfrac{Q}{4\pi \varepsilon_0}[\dfrac{1}{b}-\dfrac{3}{b}]$$$$V'_A-V'_B=\dfrac{Q}{4\pi \varepsilon_0}[\dfrac{1}{a}-\dfrac{1}{b}]=V$$The correct option is C.PhysicsNCERTStandard XII

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