Two spheres of radii 3 cm and 1 cm are plac | vedclass

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Question

(A) Let the radii of the spheres be $R_1 = 3 \ cm = 3 	imes 10^{-2} \ m$ and $R_2 = 1 \ cm = 1 	imes 10^{-2} \ m$. The potential $V$ of each sphere

Vedclass · Accepted Answer

$\left( \frac{1}{3} ight) 	imes 10^{-9} \ N$

Vedclass · Answer

(A) Let the radii of the spheres be $R_1 = 3 \ cm = 3 	imes 10^{-2} \ m$ and $R_2 = 1 \ cm = 1 	imes 10^{-2} \ m$. The potential $V$ of each sphere is $10 \ V$.The charge $Q$ on a sphere of radius $R$ at potential $V$ is given by $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{R}$,so $Q = 4\pi\epsilon_0 R V$.For the first sphere: $Q_1 = \frac{R_1 V}{k} = \frac{3 	imes 10^{-2} 	imes 10}{9 	imes 10^9} = \frac{1}{3} 	imes 10^{-10} \ C$.For the second sphere: $Q_2 = \frac{R_2 V}{k} = \frac{1 	imes 10^{-2} 	imes 10}{9 	imes 10^9} = \frac{1}{9} 	imes 10^{-10} \ C$.The distance between the centers is $d = 10 \ cm = 0.1 \ m$.The electrostatic force $F$ is given by Coulomb's Law: $F = k \frac{Q_1 Q_2}{d^2}$.$F = (9 	imes 10^9) 	imes \frac{(\frac{1}{3} 	imes 10^{-10}) 	imes (\frac{1}{9} 	imes 10^{-10})}{(0.1)^2} = (9 	imes 10^9) 	imes \frac{\frac{1}{27} 	imes 10^{-20}}{10^{-2}} = \frac{9}{27} 	imes 10^9 	imes 10^{-18} = \frac{1}{3} 	imes 10^{-9} \ N$.

Two spheres of radii $3 \ cm$ and $1 \ cm$ are placed at a distance of $10 \ cm$ from each other in a vacuum. If each sphere is charged to a potential of $10 \ V$,the force of repulsion between them is:

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