Two point charges $$+8q$$ and $$-2q$$ are located at $$x = 0$$ and $$x = L$$ respectively. The location of a point on the $$x$$ axis at which the net electric field due to these two point charges is zero is

Let the position of the point on the $$x$$-axis where the net electric field is zero be $$x$$.

The magnitude of the electric field due to a point charge is given by the formula:

$$E = \frac{k|q|}{r^2}$$

Since the two charges, $$+8q$$ and $$-2q$$, have opposite signs, the point where their electric fields cancel out must lie outside the region between them. To balance the fields, the point must also be closer to the charge with the smaller magnitude, which is $$-2q$$ located at $$x=L$$. Therefore, the null point must lie at $$x > L$$.

At this null point, the magnitude of the electric field due to the $$+8q$$ charge ($$E_1$$) must equal the magnitude of the electric field due to the $$-2q$$ charge ($$E_2$$):

$$E_1 = E_2$$

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

Dividing both sides by $$2kq$$ to simplify:

$$\frac{4}{x^2} = \frac{1}{(x - L)^2}$$

Taking the square root of both sides (since both $$x$$ and $$x-L$$ are positive distances):

$$\frac{2}{x} = \frac{1}{x - L}$$

Cross-multiplying to solve for $$x$$:

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

$$2x - 2L = x$$

$$x = 2L$$

Therefore, the location on the $$x$$-axis at which the net electric field is zero is $$x = 2L$$.