A sphere A of radius R has a charge Q on it. | vedclass

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(C) Let the distance between the centers of sphere $A$ and sphere $B$ be $r$. The electric field at point $B$ due to sphere $A$ is given by $E = \frac

Vedclass · Accepted Answer

$12E$

Vedclass · Answer

(C) Let the distance between the centers of sphere $A$ and sphere $B$ be $r$. The electric field at point $B$ due to sphere $A$ is given by $E = \frac{KQ}{r^2}$.At the midpoint between $A$ and $B$,the distance from each center is $d = \frac{r}{2}$.The electric field due to sphere $A$ (charge $Q$) at the midpoint is $E_A = \frac{KQ}{(r/2)^2} = \frac{4KQ}{r^2} = 4E$.The electric field due to sphere $B$ (charge $-2Q$) at the midpoint is $E_B = \frac{K|-2Q|}{(r/2)^2} = \frac{8KQ}{r^2} = 8E$.Since both fields are directed in the same direction (away from $A$ and towards $B$),the total electric field magnitude is $E_{total} = E_A + E_B = 4E + 8E = 12E$.

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