A nonconducting sphere of radius `R = 5 cm` has its center at the origin O of the coordinate system as shown in (Fig. 3.112). It has two spherical cavities of radius `r = 1 cm`, whose centers are at `0,3 cm` and `0,-3 cm`, respectively, and solid material of the sphere has uniform positive charge density `rho = 1 // pi mu Cm^-3`. Calculate the electric potential at point `P (4 cm, 0)`.
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Answer

`(43)/(45 pi epsilon_0) xx 10^-10` Volt

Explanation

Charge on the sphere (including cavities)
`Q = (4)/(3) pi R^3 rho = (4)/(3) pi (5 xx 10^-2)^3 (1)/(pi) xx 10^-6 = (500)/(3) xx 10^-12 C`
Charge in a volume equal to that of cavity
`q = (4)/(3) pi R^3 rho = (4)/(3) pi (1 xx 10^-2)^3 (1)/(pi) xx 10^-6 = (4)/(3) xx 10^-12 C`
Potential at P
`V_p = V_("whole sphere") - 2 V_("cavity")`
=`(k Q)/(2 R)[3 - (x^2)/(R^2)] - 2(k q)/(5 xx 10^-2)`
=`(9 xx 10^9 xx (500//3) xx 10^-12)/(2 xx 5 xx 10^-2)[ 3 - ((4)/(5))^2]`
`-(2 xx 9 xx 10^9 xx (4//3) xx 10^-12)/(5 xx 10^-2)`
=`(873)/(25) = 34.92 V`.