Question

Two point charges $Q_1$ and $Q_2$ are positioned at points $1$ and $2$, The field intensity to the right of the charge $Q_2$ on the line that passes through the two charges varies according to a law that is represented schematically in the figure.

The field intensity is assumed to be positive if its direction coincides with the positive direction on the $x-axis$.The distance between the charges is $l$. Find the sign of each charge.

• A. $Q_2$ is positive and $Q_1$ is negative
• B. $Q_2$ is positive and $Q_1$ is positive
• C. $Q_2$ is negative and $Q_1$ is negative
• D. $Q_2$ is negative and $Q_1$ is positive

Answer 1

The correct option is D $Q_2$ is negative and $Q_1$ is positive

$E=\frac{Q}{4\pi {ϵ}_{o}r}$

Since, field intensity at point 2 that is at location of $Q_2$ is $-\infty$.

Hence, $Q_2$ must be negative and only then for $r=0$, field at 2 will be $-\infty$.

Now at some distance right of point 2, field is 0.

At that point $E=\frac{Q_1}{4\pi {ϵ}_o(l+a)}+\frac{Q_2}{4\pi {ϵ}_{o}a}=0$

Since, $Q_2$ is negative and hence $Q_1$ must be positive for $E$ to be $0$..

Answer-(D)