Question

two charges 3 x 10-8 C and other is -2 x 10-8C are located 15 cm apart. at what point on the line joining tge
two charges is the electric potential zero? take the potential at infinity to be zero.​

Answer 1

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

• $q1 = 3 × {10}^{-8}$

• $q2 = -2 x {10}^{-8}$

• Distance(d) = 15 cm = 0.15m

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

• 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,

$\large{=> \frac{3 × {10}^{-8}}{x} \:=\: - \frac{-2 x {10}^{-8}}{0.15-x}}$

$\large{=> \frac{3}{x} \:=\: \frac{2}{0.15-x}}$

On Cross Multiplication we get,

$=> 2x\: =\: 0.50\: - \:3x$

$=> 5x\: =\: 0.50$

$=> x \:= \:0.10$

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

Thus, the Electric Potential is zero at 10 cm at right side of charge q(1) or (15-10) 5 cm at left side of charge q(2).

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