Two spherical conductors A and B of radii R_1 | vedclass

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(B) When two conductors are connected by a conducting wire,they reach the same electric potential $V$. Let $Q_1$ and $Q_2$ be the charges on spheres $

Vedclass · Accepted Answer

$R_2 / R_1$

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(B) When two conductors are connected by a conducting wire,they reach the same electric potential $V$. Let $Q_1$ and $Q_2$ be the charges on spheres $A$ and $B$ respectively. The potential on the surface of a sphere is given by $V = k \cdot (Q/R)$. Since the potentials are equal,$k \cdot (Q_1 / R_1) = k \cdot (Q_2 / R_2)$,which implies $Q_1 / Q_2 = R_1 / R_2$. The electric field $E$ at the surface of a sphere is given by $E = k \cdot (Q / R^2)$. Therefore,the ratio of the electric fields is $E_1 / E_2 = (Q_1 / R_1^2) / (Q_2 / R_2^2) = (Q_1 / Q_2) \cdot (R_2^2 / R_1^2)$. Substituting $Q_1 / Q_2 = R_1 / R_2$,we get $E_1 / E_2 = (R_1 / R_2) \cdot (R_2^2 / R_1^2) = R_2 / R_1$.

Two spherical conductors $A$ and $B$ of radii $R_1$ and $R_2$ are charged to the same potential and placed at a distance $d$ apart. If these spheres are connected by a conducting wire,what is the ratio of the magnitudes of the electric fields at the surfaces of $A$ and $B$ in the equilibrium state?

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