The point charges +q, -q, -q, +q, +Q and -q a | vedclass

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

Question

(A) Let $r$ be the distance from each vertex to the centre $O$. The electric field at $O$ due to a charge $q$ at a distance $r$ is $E = \frac{kq}{r^2}

Vedclass · Accepted Answer

$\frac{+q}{3}$

Vedclass · Answer

(A) Let $r$ be the distance from each vertex to the centre $O$. The electric field at $O$ due to a charge $q$ at a distance $r$ is $E = \frac{kq}{r^2}$.$1$. The charges at $A (+q)$ and $D (+q)$ produce electric fields at $O$ that are equal in magnitude but opposite in direction. Thus, they cancel each other.$2$. The charges at $F (-q)$ and $C (-q)$ produce electric fields at $O$ that are equal in magnitude but opposite in direction. Thus, they cancel each other.$3$. The only remaining charge from the set ${A, B, C, D, F}$ is the charge $-q$ at vertex $B$. The electric field at $O$ due to this charge is $E_{net} = \frac{kq}{r^2}$ directed towards $B$.$4$. The electric field at $O$ due to charge $+Q$ at $E$ is $E_Q = \frac{kQ}{r^2}$ directed away from $E$.$5$. According to the problem, $E_{net} = 3 E_Q$.$6$. Substituting the values: $\frac{kq}{r^2} = 3 \left( \frac{kQ}{r^2} \right)$.$7$. Solving for $Q$, we get $Q = \frac{q}{3}$.

Similar Questions

Considering a group of positive charges,which of the following statements is correct?

Four positive point charges $+q$ are kept at the four corners of a square of side $l$. The net electric field at the midpoint of any one side of the square is (take $\frac{1}{4 \pi \epsilon_0}=k$ ).

The electric field due to a point charge $2q$ at a distance $r$ is $E$. Now, if charge $q$ is uniformly distributed over a thin spherical shell of radius $R$, the electric field at a distance $\frac{r}{2}$ $(r \gg R)$ from the center of the thin spherical shell is $E'=$ . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module