Question

A few electric field lines for a system of two charges $Q_1$ and $Q_2$ fixed at two different points on the $x-$axis are shown in the figure. These lines suggest that

• A. $\left|Q_1\right|>\left|Q_2\right|$
• B. $\left|Q_1\right|<\left|Q_2\right|$
• C. at a finite distance to the left of $Q_1$ the electric field is zero
• D. at a finite distance to the right of $Q_2$ the electric field is zero

Answer 1

Answer: (A), (D)

$\left|Q_1\right|>\left|Q_2\right|$
at a finite distance to the right of $Q_2$ the electric field is zero

As electric field lines are originating from $Q_1$ and terminating at $Q_2$

$\Rightarrow Q_1$ is +ve and $Q_2$ is (-)ve.

Density of field lines $\alpha$ strength of electric field $\alpha$ $\left|Q\right|$

So, $\left|Q_1\right|>\left|Q_2\right|$ (A)

$\Rightarrow E_1=$ Electric field at P due to $Q_1$

$E_2=$ Electric field at P due to $Q_2$

For $E_p=0,E_1-E_2=0$

$\Rightarrow E_1-E_2=0$

$\Rightarrow E_1=E_2$

$\Rightarrow \frac{k{Q}_1}{{(r+d)}^2}=\frac{k{Q}_2}{r^2}$

$\Rightarrow \frac{Q_1}{Q_2}=\frac{{(r+d)}^2}{r^2}$

As $\left|Q_1\right|>\left|Q_2\right|,$ so, such ${}^{\prime }{r}^{\prime }$ exists.

Hence, both option (A) and (D) are correct.