91Ó°ÊÓ

We are given two particles having charges Q₁ +0.500 nC and q₂ = 8.00 nC are separated by a distance of 1.20 m.

Each charge has an electric field, and as both of them have a positive charge, the electric field from each charge is away from the charge. Also, the total electric field will be zero when the two electric fields have the same magnitude but in opposite direction.

As shown in the figure below, the region between the two charges has the condition that the two electric fields

are in the opposite direction, and let us consider that point o is the point at which the total electric field is zero, where the distance between qi and o is x while between q₂and o is (1.20 m – x).

Now calculate the value of x to determine the required, where the two electric fields at the point have the same magnitude

$\begin{array}{r}E1=E2\\ \frac{k\left|q_1\right|}{x^2}=\frac{k\left|q2\right|}{(1.20−\mathrm{x}{)}^2}\\ \frac{\left|q_1\right|}{x^2}=\frac{\left|q2\right|}{(1.20−\mathrm{x}{)}^2}\\ \frac{(1.20−\mathrm{x})}{x}=\sqrt{\frac{q_2}{q_1}}\\ \frac{1.20}{x}=\sqrt{\frac{q_2}{q_1}}\end{array}$ ……… (1)

Now substitute the given values in equation(1)

$\begin{array}{r}x=\frac{1.20}{\sqrt{\frac{8.00nC}{0.500nC}}+1}\\ =0.24m\end{array}$

Hence the point at which the total electric field between two charges is equal to zero is at distance 0.24m from q₁.