Question 796: Electrostats & Capacitance

Question

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

Accepted Answer

To determine the point on the x-axis where the net electric field due to two charges $ +8q $ and $ -2q $ is zero, let's denote this point as $ A $. Suppose $ A $ is located at a distance $ r $ from the charge $ -2q $ which is at $ x = L $, and thus at a distance $ L + r $ from the charge $ +8q $ located at $ x = 0 $.
To find the condition where the electric fields from both charges cancel each other, set the magnitudes of the electric fields equal to each other:
$ E_1 = E_2 $
Given by:
$ \frac{K \cdot 8q}{(L + r)^2} = \frac{K \cdot 2q}{r^2} $
Simplifying, we have:
$ \frac{4}{(L + r)^2} = \frac{1}{r^2} $
Solving the equation, we find:
$ 2r = L + r $
$ r = L $
Therefore, the position on the x-axis from the origin where the electric fields cancel is:
$ L + r = L + L = 2L $
Hence, the point is located at $ 2L $ from the origin.

Answer

$\frac{\mathrm{L}}{4}$

Answer

$\mathrm{4 L}$

Answer

$\mathrm{8 L}$

Answer

$\mathrm{2 L}$

Electrostats & Capacitance

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

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