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NCERT Class XII Chapter
Electrostatic Potential and Capacitance
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Question : 29 of 37
Marks: +1, -0
A spherical capacitor consists of two concentric spherical conductors held in position by suitable insulating supports. Show that the capacitance of a spherical capacitor is given by C = 4πε0r1r2r1−r2\frac{4\pi\varepsilon_0 r_1 r_2}{r_1 - r_2}
Where r1r_1 and r2r_2 are the radii of outer and inner spheres, respectively.
Solution:  
Let + Q be the charge on outer spherical shell A of radius r1r_1 and + Q be the charge on inner spherical shell B of radius r2r_2. Then electric potential on shell A is
VAV_A = VAA+VABV_{AA} + V_{AB} = 14πε0\frac{1}{4\pi\varepsilon_0} [+Qr1−Qr1]\left[+\frac{Q}{r_1}-\frac{Q}{r_1}\right] or VAV_A = 0
and electric potential on shell B is
VBV_B = VBA+VBBV_{BA} + V_{BB} = 14πε0[Qr1−Qr2]\frac{1}{4\pi\varepsilon_0} \left[ \frac{Q}{r_1} - \frac{Q}{r_2} \right]
or VBV_B = Q4πε0\frac{Q}{4\pi\varepsilon_0} [r2−r1r1r2]\left[ \frac{r_2 - r_1}{r_1 r_2} \right]
So, the potential difference between the two spherical shells A and B is
V = VA−VBV_A - V_B = 0 - Q4πε0[r2−r1r1r2]\frac{Q}{4\pi\varepsilon_0} \left[ \frac{r_2 - r_1}{r_1 r_2} \right]
or V = Q4πε0[r2−r1r1r2]\frac{Q}{4\pi\varepsilon_0} \left[ \frac{r_2 - r_1}{r_1 r_2} \right] or QV\frac{Q}{V} = 4πε0r1r2r1r2\frac{4\pi\varepsilon_0 r_1 r_2}{r_1 r_2} or QV\frac{Q}{V} = 4πε0r1r2r1−r2\frac{4\pi\varepsilon_0 r_1 r_2}{r_1 - r_2} or C = 4πε0r1r2r1−r2\frac{4\pi\varepsilon_0 r_1 r_2}{r_1 - r_2}
This gives the capacitance of the spherical capacitor.
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