The number of electrons to be put on a spheri | vedclass

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Question

(C) The electric field $E$ just above the surface of a spherical conductor is given by $E = \frac{kQ}{R^2}$,where $k = 9 	imes 10^9 \ N \cdot m^2/C^2

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(C) The electric field $E$ just above the surface of a spherical conductor is given by $E = \frac{kQ}{R^2}$,where $k = 9 	imes 10^9 \ N \cdot m^2/C^2$,$Q = ne$ is the total charge,and $R$ is the radius.Given: $E = 0.036 \ N/C = 3.6 	imes 10^{-2} \ N/C$,$R = 0.1 \ m = 10^{-1} \ m$,and $e = 1.6 	imes 10^{-19} \ C$.Substituting the values: $3.6 	imes 10^{-2} = \frac{9 	imes 10^9 	imes n 	imes 1.6 	imes 10^{-19}}{(10^{-1})^2}$.$3.6 	imes 10^{-2} = \frac{14.4 	imes 10^{-10} 	imes n}{10^{-2}}$.$3.6 	imes 10^{-2} = 14.4 	imes 10^{-8} 	imes n$.$n = \frac{3.6 	imes 10^{-2}}{14.4 	imes 10^{-8}} = \frac{3.6}{14.4} 	imes 10^6 = 0.25 	imes 10^6 = 2.5 	imes 10^5$.Since the question asks for the value in terms of $10^4$,we write $n = 25 	imes 10^4$.

The number of electrons to be put on a spherical conductor of radius $0.1 \ m$ to produce an electric field of $0.036 \ N/C$ just above its surface is $...... \times 10^4$.

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