CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
loader
B
4V
loader
C
V
loader
D
2V
loader

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.


1451511_11243_ans_253bbf8593de44168a9452edf16a80e3.jpeg

Physics
NCERT
Standard XII

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
Same exercise questions
View More


similar_icon
People also searched for
View More



footer-image