Let's work through the problem step by step.

Identify the charges and their positions:

A charge of $$+8q$$ is at the origin ($$x=0$$).

A charge of $$-2q$$ is at $$x=L$$.

Determine the region where the net electric field could be zero:

If we look between the charges (i.e., $$0 < x < L$$), the electric fields due to both charges will point in the same direction along the $$x$$‑axis. Hence, they cannot cancel.

For a point to the right of both charges (i.e., $$x > L$$), the field due to the positive charge points to the right (away from it), while the field from the negative charge points to the left (toward it). This setup allows for cancellation.

Write expressions for the electric fields at a point $$x$$ (where $$x > L$$):

The electric field due to $$+8q$$ at a distance $$x$$ is:

$$E_1 = k\frac{8q}{x^2}$$ (pointing right).

The electric field due to $$-2q$$ at a distance $$(x-L)$$ is:

$$E_2 = k\frac{2q}{(x-L)^2}$$ (pointing left, because the field due to a negative charge is towards the charge).

Set up the condition for net electric field being zero:

Since the two fields are in opposite directions, they cancel when their magnitudes are equal:

$$k\frac{8q}{x^2} = k\frac{2q}{(x-L)^2}$$

Simplify the equation:

Cancel $$k$$ and $$q$$ from both sides:

$$\frac{8}{x^2} = \frac{2}{(x-L)^2}$$

Cross-multiply:

$$8(x-L)^2 = 2x^2$$

Divide both sides by 2:

$$4(x-L)^2 = x^2$$

Solve for $$x$$:

Take the square root of both sides (note that since $$x > L$$, $$x-L$$ is positive):

$$x = 2(x-L)$$

Distribute and solve:

$$x = 2x - 2L$$

$$2L = 2x - x$$

$$x = 2L$$

Conclusion:

The point at which the net electric field is zero is $$x=2L$$ from the origin.

Therefore, the correct answer is Option B: $$2L$$.

Electrostatics

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