1. Given charges and positions

• Charge $+8q$ is at $x=0$
• Charge $-2q$ is at $x=L$

We need the point on the $x$-axis where the net electric field is zero.

2. Consider different regions on the $x$-axis

The electric field due to a point charge depends on direction as well as magnitude, so we check region-wise.

Region I: $x<0$

Let the point be at $x=-a$ where $a>0$.

• Field due to $+8q$ points away from positive charge, so at $x<0$ it is toward left.
• Field due to $-2q$ points toward the negative charge, so at $x<0$ it is toward right.

So the two fields are opposite in direction, hence cancellation is possible.

Magnitudes: $E_1 = k\frac{8q}{a^2}, \qquad E_2 = k\frac{2q}{(L+a)^2}$

For cancellation: $k\frac{8q}{a^2} = k\frac{2q}{(L+a)^2}$ $\frac{8}{a^2} = \frac{2}{(L+a)^2}$ $4(L+a)^2 = a^2$ $2(L+a)=a$ $2L+2a=a$ $a=-2L$

This is impossible since $a>0$. Hence no solution in $x<0$.

Region II: $0<x<L$

Let the point be between the charges.

• Field due to $+8q$ points away from it, i.e. toward right.
• Field due to $-2q$ points toward it, also toward right.

Both fields are in the same direction, so they cannot cancel.

Hence no solution in $0<x<L$.

Region III: $x>L$

Let the point be at $x=L+a$ where $a>0$.

• Distance from $+8q$: $L+a$
• Distance from $-2q$: $a$

Directions:

• Field due to $+8q$ is toward right.
• Field due to $-2q$ is toward left.

So cancellation is possible.

Magnitudes: $E_1 = k\frac{8q}{(L+a)^2}, \qquad E_2 = k\frac{2q}{a^2}$

Set equal for net field zero: $k\frac{8q}{(L+a)^2} = k\frac{2q}{a^2}$ $\frac{8}{(L+a)^2} = \frac{2}{a^2}$ $4a^2 = (L+a)^2$ $2a = L+a$ $a=L$

Therefore, $x=L+a=L+L=2L$

3. Check options

• A: $\dfrac{L}{4}$ ❌
• B: $2L$ ✅
• C: $4L$ ❌
• D: $8L$ ❌

4. Final answer

The net electric field is zero at $\boxed{2L}$ which corresponds to Option B.

5. Comparison with stored correct answer

Stored correct answer: B

My derived answer: B

So they agree.