Question

two charges 5×10^(-8) C and -3×10^(-8) are located 16cm apart. at what points on the line joining the two charges is the electric potential zero ? take the potential at infinity to be zero.​

Answer 1

$\sf \: {\blue{ \underline{ \underline{Answer ~:}}}}$

Given :

$\sf \: q_1 \: = \: 5 \times {10}^ {- 8}$

$\sf \: q_2 = - 3 \times {10}^{ - 8}$

Distance(d) = 16cm = 0.16m

To Find :

Position where the electric potential is 0.

Solution :

$\setlength{\unitlength}{1cm}\thicklines\begin{picture}(10,6)\put(1,1){\vector(-1,0){3}} \put(1,1){\vector(1,0){3}}\put(-1.5,1){\circle*{0.1}}\put(1.5,1){\circle*{0.1}}\put(3.5,1){\circle*{0.1}}\put(-1.5,2){\line(0,1){0.3}}\put(-1.5,2.15){\line(1,0){3}}\put(1.5,2){\line(0,1){0.3}}\put(-1.5,-0.5){\line(0,1){0.3}}\put(-1.5,-0.35){\line(1,0){5}}\put(3.5,-0.5){\line(0,1){0.3}}\put(-1.6,0.5){\sf A}\put(1.35,0.5){\sf B}\put(3.4,0.5){\sf C}\put(-0.2,1.7){\sf x\;cm}\put(0.2,0){\sf d = 16\;cm}\put(-1.6,1.4){ \sf q_1 }\put(3.4,1.4){ \sf q_2 }\end{picture}$

● Let C be the point from 'x cm' to the right of charge q(1) where electric potential is zero. (as shown in the fig.)

According to Question,

$\boxed{ \sf \: V = \: \frac{kq_1}{r_1} + \frac{kq_2}{r_2} }$

$\implies \: \sf \: V = \: \frac{kq_1}{x} + \frac{kq_2}{(d - x)}$

When, Electric Potential is zero.

$\implies \: \sf \: 0 = \: \frac{kq_1}{x} + \frac{kq_2}{(d - x)}$

$\implies \: \sf \: \frac{kq_1}{x} = - \frac{kq_2}{(d - x)}$

$\implies \: \sf \: \frac{ \cancel{k}q_1}{x} = - \frac{ \cancel{k}q_2}{(d - x)}$

$\implies \: \sf \: \frac{q_1}{x} = - \frac{q_2}{(d - x)}$

On putting the values,

$\implies \: \sf \: \frac{5 \times {10}^{ - 8} }{x} = \frac{ - ( - 3 \times {10}^{- 8})}{0.16 \: - \: x}$

$\implies \: \sf \: \frac{5 \times \cancel{{10}^{- 8}}}{x} = \frac{ 3 \times \cancel{{10}^{- 8}}}{0.16 \: - \: x}$

On Cross Multiplication, we get

$\implies \sf \: \: 3x = \: 0.80 \: - \: 5x$

$\implies \sf \: \: 8x = \: 0.80$

$\implies \: \sf x \: = 0.10 \:$

$\boxed{ \sf \: \therefore \: x =0 .10 \: m \: or \: 10 \: cm}$

Thus, the Electric Potential is zero at 10cm at right side of charge q(1) or (16 - 10) = 6cm at left side of charge q(2).

Answer 2

Answer:

$\sf{q_{1} = 5 \times 10^{-8}}$

$\sf{q_{2} = -3 \times 10^{-8}}$

$\implies\sf{ \frac{5 \times 10^{-8}}{x} = \frac{ -(-3 \times 10^{-8}}{0.16 -x} }$

$\longrightarrow\sf{ \frac{5 \times \cancel{10^{-8}}}{x} = \frac{ -(-3 \times \cancel{10^{-8}}}{0.16 -x} }$

$\implies\sf{ 3x = 0.80 - 5x}$

$\implies\sf{ 3x + 5x = 0.80}$

$\implies\sf{ 8x = 0.80}$

$\implies\bf\green{\fbox{ x = 0.10}}$

Hope it will be helpful ✌️