Question

Two charges $+16q$ and $q$ are set on the line shown, where the position $x=0$ is the midpoint between the two charges. The two charges create an electric field in the region around them.
At which x-coordinate is the electric field equal to zero?

• A. $x=+10m$
• B. $x=+4m$
• C. $x=-4m$
• D. $x=0m$
• E. $x=-10m$

Answer 1

The correct option is A $x=+10m$

Consider the region between $x=-\infty$ and $x=-6$.

The electric field due to $16q$ is towards left and that due to $-q$ is towards right.

Electric field at a general point $x=d(<-6)$ is $E=\frac{1}{4\pi {ϵ}_0}(-\frac{16q}{(-6-d{)}^2}+\frac{q}{(6-d{)}^2})$

If $E=0$, then $\frac{16q}{(-6-d{)}^2}=\frac{q}{(6-d{)}^2}\Rightarrow \frac{(6+d{)}^2}{(6-d{)}^2}=16$

But, if $d<-6$, then $|6+d|<|6-d|$. Hence, no solution exists.

Consider the region between $x=-6$ and $x=+6$

The electric field due to $+16q$ is towards right, and that due to $-q$ is towards right.

Hence, there is no point in$(-6,6)$ with zero electric field.

Consider the region between $x=+6$ and $x=+\infty$

The electric field due to $+16q$ is towards right and $-q$ is towards left.

Electric field at a general point $x=d(>6)$ is given by $E=\frac{1}{4\pi {ϵ}_0}(+\frac{16q}{(d+6{)}^2}-\frac{q}{(d-6{)}^2})$

If $E=0$, then $\frac{16q}{(d+6{)}^2}=\frac{q}{(d-6{)}^2}\Rightarrow \frac{d+6}{d-6}=\pm 4\Rightarrow d=10\text{or}d=3.6$

But, $d>6$ and hence $d=10\text{m}$

Thus at $x=10m$, electric field is zero.