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

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

Two point charges $Q_1$ and $Q_2$ are fixed at $x = 0$ and $x = a$. Assuming that the electric field strength is positive in the direction coinciding with the positive direction of $x$,which of the following options is correct?

Four point charges are placed at the vertices of a regular hexagon as shown in the figure below. What will be the net electric field at the centre?

Two point charges $-q$ and $+q/2$ are situated at the origin $(0, 0, 0)$ and at the point $(a, 0, 0)$ respectively. The point along the $X$-axis where the electric field vanishes is:

Is electric field a scalar or a vector quantity? Why?

Vedclass Test Series

Exam Paper Generator

Online Exam Module