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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $E_{total} = E_A + E_B = 4E + 8E = 12E$.

Similar Questions

Two charges $Q$ and $-3Q$ are placed at a certain distance $x$. If the electric field at the position of charge $Q$ due to charge $-3Q$ is $E$,then what is the electric field at the position of charge $-3Q$ due to charge $Q$?

$A$ ring of radius $R$ is charged uniformly with a charge $+Q$. The electric field at a point on its axis at a distance $r$ from any point on the ring will be

Four charges $q, 2q, -4q$ and $2q$ are placed in order at the four corners of a square of side $b$. The net field at the centre of the square is

The variation of electric field along the $Z$-axis due to a uniformly charged circular ring of radius '$a$' in the $XY$ plane is shown in the figure. The value of coordinate $M$ will be

$A$ uniform electric field of $500 \ Vm^{-1}$ is directed at $30^{\circ}$ with the positive $X$-axis as shown in the figure. The potential difference $(V_B - V_A)$ if $OA = 3 \ m$ and $OB = 5 \ m$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module