Question

Two charges 5×10-8 C and -3×10-8 C are located 16 cm apart. At what point(s) on the line joining the
two charges is the electric potential zero? Take the potential at infinity to be zero.

Answer 1

Explanation:

at x = 32/5

the line joinining of two charge = 0

Answer 2

$\huge\green{\underline{\underline{Given :}}}$

$q1 = 5 × {10}^{-8}$

$q2 = -3 x {10}^{-8}$

Distance(d) = 16cm = 0.16m

$\huge\green{\underline{\underline{To\:Find :}}}$

Position where the electric potential is 0.

$\huge\green{\underline{\underline{Solution :}}}$

$\huge{•}$ 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,

$\pink{\boxed{{V \:=\: }\frac{kq1}{r1}+\frac{kq2}{r2}}}$

$=>\large{ V \:=\: \frac{kq1}{x}+\frac{kq2}{d-x}}$

When, Electric Potential is zero,

$=> \large{0 \:=\: \frac{kq1}{x}+\frac{kq2}{d-x}}$

$\large{=> \frac{kq1}{x} \:=\: - \frac{kq2}{d-x}}$

$\large{=> \frac{q1}{x} \:=\: - \frac{q2}{d-x}}$

On putting the values,

$\large{=> \frac{5 × {10}^{-8}}{x} \:=\: - \frac{-3 x {10}^{-8}}{0.16-x}}$

$\large{=> \frac{5}{x} \:=\: \frac{3}{0.16-x}}$

On Cross Multiplication we get,

$=> 3x\: =\: 0.80\: - \:5x$

$=> 8x\: =\: 0.80$

$=> x \:= \:- 0.10$

$\large\pink{\fbox{=>\:x\: =\: 0.10\: m\: or\: 10\: cm}}$

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

HOPE IT HELPS

PLEASE MARK ME BRAINLIEST ☺️