Question

Charges $2q$ and $-q$ are placed at $(a,0)$ and $(-a,0)$ as shown in the figure. The coordinates of the point at which electric field intensity is zero is $(x,0)$. Then:

• A. $x>-a$
• B. $-a<x<a$
• C. $0<x<a$
• D. $x<-a$

Answer 1

The correct option is D $x<-a$

We know that, electric field due to a point charge is given by,
$E=\frac{kQ}{r^2}[E\propto Q\text{and}E\propto \frac{1}{r^2}]$


For point $O(-a<x<a)$, direction of electric field due to both charges will be same, so net electric field cannot be zero for $-a<x<a$.

At point $A(x>a)$, direction of electric field due to both charges are in opposite directions.
But, magnitude of electric field due to $+2q$ charge will be greater as magnitude of charge is greater than $-q$ and also distance is less than that of $-q$.
Therefore, net electric field cannot be zero for $x>a$ as well.

At point $B(x<-a)$, direction of electric field due to both charges are in opposite directions and also magnitude of electric field may become equal as distance is less for charge $-q$, so it may compensate for the effect of smaller value of charge.

Hence, for some point $B(x<-a)$, the net electric field will be zero.