We have an isolated conducting spherical shell of radius `10 cm`. Some positive charge is given to it so that the resulting electric field has a maximum intensity of `1.8 xx 10^6 NC^-1`. The same amount of negative charge is given to another isolated conducting spherical shell of radius `20 cm`. Now, the first shell is placed inside the second so that both are concentric as shown in (Fig. 3.154).
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The electric potential at any point inside the first shell is.

Options

A. `18 xx 10^4 V`
B. `9 xx 10^4 V`
C. `4.5 xx 10^4 V`
D. `1.8 xx 10^4 V`

Answer

B

Explanation

`1.8 xx 10^6 = (9 xx 10^9 xx q_1)/((0.1)^2) or q_1 = 2 xx 10^-6 C`
`V = 9 xx 10^9 xx 2 xx 10^-10 [(1)/(0.1) - (1)/(0.2)] = 9 xx 10^4 V`
`E = (k q_1)/(r^2) = (9 xx 10^9 xx 2 xx10^-6)/((0.20)^2)= 4.5 xx 10^5 NC^-1`
Interaction energy is
`U = (k q)/(0.2) (-q) =(9 xx 10^9 xx (2 xx 10^-6)^2)/(0.2) = -0.18 J`
Self - energies are
`U_1 = q_(2)/(8 pi epsilon_0 r_1) = (9 xx 10^9 xx (2 xx 10^-6)^2)/(2 xx 0.1)= 0.18 J`
`U_(2)=q_(2)/(8pi epsilon_(0) r_(2)) =(9xx10^(9)xx(2xx10^(-6))^(2))/(2xx0.2)=0.09 J`
Total energy is `U + U_1 + U_2 = 0.09 J`
Both the charges will get neutralized and there will be no charge left on any sphere. So no energy will be left in the system. It means whole amount of energy will convert into heat.
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