Two neutral conducting spheres of diameters 8 | vedclass

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(A) Let the radii of the spheres be $R_1 = 4 \ cm = 0.04 \ m$ and $R_2 = 1 \ cm = 0.01 \ m$. The distance between centers is $d = 0.15 \ m$.When conne

Vedclass · Accepted Answer

$10.8 	imes 10^3$

Vedclass · Answer

(A) Let the radii of the spheres be $R_1 = 4 \ cm = 0.04 \ m$ and $R_2 = 1 \ cm = 0.01 \ m$. The distance between centers is $d = 0.15 \ m$.When connected by a wire,the spheres reach the same potential $V = \frac{k q_1}{R_1} = \frac{k q_2}{R_2}$. Thus,$q_1/q_2 = R_1/R_2 = 4/1$.Given $q_1 + q_2 = 100 \ nC$,we get $q_1 = 80 \ nC$ and $q_2 = 20 \ nC$.The potential is $V = \frac{9 	imes 10^9 	imes 80 	imes 10^{-9}}{0.04} = 18000 \ V$.Let the point where the electric field is zero be at distance $x$ from the center of the first sphere. The field is zero when $\frac{k q_1}{x^2} = \frac{k q_2}{(d-x)^2}$.Taking the square root: $\frac{\sqrt{80}}{x} = \frac{\sqrt{20}}{d-x} \implies \frac{2\sqrt{20}}{x} = \frac{\sqrt{20}}{0.15-x}$.$2(0.15 - x) = x \implies 0.3 = 3x \implies x = 0.1 \ m$.The potential at this point is $V_P = \frac{k q_1}{x} + \frac{k q_2}{d-x} = \frac{9 	imes 10^9 	imes 80 	imes 10^{-9}}{0.1} + \frac{9 	imes 10^9 	imes 20 	imes 10^{-9}}{0.05} = 7200 + 3600 = 10800 \ V = 10.8 	imes 10^3 \ V$.

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