Question

A solid metal sphere of radius a = 2.5 cm has a net charge Qin = - 3 nC (1 nC = 10-9C). The sphere is surrounded by a concentric conducting spherical shell of inner radius b = 6 cm and outer radius c = 9 cm. The shell has a net charge Qout = + 2 nC. What is V0, the electric potential at the center of the metal sphere, given the potential at infinity is zero? V0 =

Answer 1

Answer: 630 V

Explanation: In order to solve this problem we have to consider the potential given by:

In the region 0<a<b

V(r)= -∫E*dr where E can be calculated by Gaussian law then E= k*qa/r^2

where qa=-3 nC

then

The V(r)=k*qa/r+C C is zero since V(r=∞)=0

Finally we have

V(r)= k*qa (1/r)-(1/b)

thus V(a)= k*qa (1/a)-(1/b)

Replacing the values V(a) = 630 V, as the solid metal sphere is a equipotential object thus at the center have the same value of V that in r=a ( 630 V).

Answer 2

We have that for the Question "hat is V0, the electric potential at the center of the metal sphere, given the potential at infinity is zero? V0 =" it can be said that

• V_o=-730V

From the question we are told

• A solid metal sphere of radius a = 2.5 cm has a net charge Qin = - 3 nC (1 nC = 10-9C).
• The sphere is surrounded by a concentric conducting spherical shell of inner radius b = 6 cm and outer radius c = 9 cm. The shell has a net charge Qout = + 2 nC.
• What is V0, the electric potential at the center of the metal sphere, given the potential at infinity is zero? V0 =

Voltage

Generally the equation for the Charge on outer surface  is mathematically given as

$Q_o=(2nC-3nC)\\\\Q_o=-1*10^{-9}\\\\Therefore\\\\V_o=\frac{kQ_1}{a}+\frac{kQ_2}{b}+\frac{kQ_o}{c}\\\\Hence\\\\V_o=\frac{-k3nc}{0.025}+\frac{k3nc}{0.06}+\frac{-k1nc}{0.09}$

V_o=(9x10. EB 10)(-8111x10. = —730 V

V_o=-730V

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