Question

Two point charges 5*10-19 C and 20*10--19 C are separated by a distance of 2m. Find the point on the line
joining them at which the electric field intensity is zero.

Answer 1

See the below picture related to answer

Answer 2

The point in the line where the intensity of electric field be 1.33 m from the second point charge.

Solution:

The point on the line joining the two point charges be given by,

Given: Charge $q_{1}=5 \times 10^{-19} C$

            Charge $q_{2}=20 \times 10^{-19} C$

Distance between the point charges $q_{1}$ and $q_{2}$  = 2 m

Let us assume the point at which electric field intensity is zero be at a distance from $q_{2} = x$  

And from $q_{1}$ = 2-x

As given that $\bold{E_{1}-E_{2}=0}$  

$\begin{array}{l}{E_{1}=E_{2}} \\ \\{E=\frac{K q}{r^{2}}} \\ \\{\frac{K q_{1}}{r_{1}^{2}}=\frac{K q_{2}}{r_{2}^{2}}} \\ \\{\frac{q_{1}}{r_{1}^{2}}=\frac{q_{2}}{r_{2}^{2}}}\end{array}$

$\begin{array}{l}{\frac{5 \times 10^{-19}}{(2-x)^{2}}=\frac{20 \times 10^{-19}}{x^{2}}} \\ \\{\frac{5}{(2-x)^{2}}=\frac{20}{x^{2}}} \\ \\{\frac{1}{(2-x)^{2}}=\frac{4}{x^{2}}} \\ \\{\text { Squaring both sides, } \frac{1}{(2-x)}=\frac{2}{x}}\end{array}$

x = 4 – 2x

x + 2 x = 4

3x =4

x=$\frac{4}{3}$

x = 1.33  m.