# Two-point charges + 8q and q are located at x = 0 and x = L respectively. The location of a point...

## Question:

Two-point charges + 8q and q 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

(a) L/4

(b) 2L

(c) 4 L

(d) 8 L

## Electric Field Due to Stationary Charge:

Consider a stationary particle having charge {eq}q {/eq} C kept at the position vector {eq}\overrightarrow{x} {/eq}. Then due to this particle, the electric field at a point with position vector {eq}\overrightarrow{p} {/eq} is given by $$\overrightarrow{E} = \frac{q (\overrightarrow{p} - \overrightarrow{x})}{4\pi \epsilon_0 |\overrightarrow{p}-\overrightarrow{x}|^3} $$

## Answer and Explanation:

Charge +8q is kept at x = 0, and charge +q is kept at x = L. Since both these charges are positive, therefore the electric field due to them will be outwards from their position.

For x < 0, both charges will given field in the -x direction. Hence the field for x < 0 can't be zero.

Similarly, for x > L, the field can't be zero.

Therefore, we find that for some x between 0 and L, the field will be zero.

Now, if this point does not lie on the x-axis, then the field from both these particles will have an x- and y- component. The x-components of both the fields will be opposite, but the y-component will be in the same direction. Hence, for the field to be zero, the point must lie on the x-axis.

Let the point have the coordinate (x,0)

Then the magnitude of the field due to first particle will be. {eq}E_1 = \frac{8q}{4\pi\epsilon_0 (x-0)^2} = \frac{8q}{4\pi\epsilon_0 x^2}\\ {/eq}

The magnitude of the field due to the second particle will be {eq}E_2 = \frac{q}{4\pi\epsilon_0 (L-x)^2} {/eq}

Now, when the field will be zero, both magnitudes will be equal, $$E_1 = E_2\\ \frac{8q}{4\pi\epsilon_0 x^2} = \frac{q}{4\pi\epsilon_0 (L-x)^2}\\ \frac{8}{x^2} = \frac{1}{(L-x)^2}\\ 8(L-x)^2 = x^2\\ 2\sqrt2 (L-x) = x\\ x = \frac{2\sqrt2 L}{2\sqrt2+1} $$

Therefore, at x = {eq}\frac{2\sqrt2 L}{2\sqrt2 +1} {/eq}, the net electric field will be zero.

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