Two point charges Q_1 and Q_2 are fixed at x | vedclass

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

Question

(C) $1$. For $x < 0$,the electric field $E$ is negative. Since $E$ is directed towards the negative $x$-axis near $x=0$,$Q_1$ must be negative.$2$. Fo

Vedclass · Accepted Answer

$Q_1$ is negative and $Q_2$ is positive with $|Q_1| > |Q_2|$.

Vedclass · Answer

(C) $1$. For $x $2$. For $x > a$,the electric field $E$ is positive. Since $E$ is directed towards the positive $x$-axis near $x=a$,$Q_2$ must be positive.$3$. The electric field becomes zero at some point $x > a$. This implies that the magnitude of the electric field due to $Q_1$ at that point must be equal to the magnitude of the electric field due to $Q_2$. Since the point where $E=0$ is closer to $Q_2$ than to $Q_1$,it follows that $|Q_1| > |Q_2|$.$4$. Therefore,$Q_1$ is negative,$Q_2$ is positive,and $|Q_1| > |Q_2|$. The correct option is $C$.

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?

Similar Questions

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:

Two charges, each equal to $-q$, are kept at $(-a, 0)$ and $(a, 0)$. $A$ charge $q$ is placed at the origin. If $q$ is given a small displacement $y$ along the $y$-direction, the force acting on $q$ is proportional to:

Two point charges of magnitudes $-8 \mu C$ and $+32 \mu C$ are separated by a distance of $15 \ cm$ in air. The position of the point from the $-8 \mu C$ charge at which the resultant electric field becomes zero is (in $cm$)

For the given figure,the direction of the electric field at point $A$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module