Question

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

Answer 1

two charges $q_1=5\times10^{-8}C$ and $q_2=-3\times10^{-8}C$ are located 16cm or 0.16m apart.

Let electric potential becomes zero at r m distance from $q_1$ and then electrical potential becomes zero at (0.16 - r) m from $q_2$.

so, electric potential at that point = potential due to $q_1$+ potential due to $q_2$

or, 0 = $\frac{kq_1}{r}+\frac{kq_2}{(0.16-r)}$

or, 0 = k [ 5 × 10^-8/r + (-3 × 10^-8)/(0.16 - r)]

or, 5/r = 3/(0.16 - r)

or, 5(0.16 - r) = 3r

or, 0.8 - 5r = 3r

or, r = 0.1m

hence, electric potential is zero at 0.1 m or 10cm from $q_1$ and 0.06m or 6cm from $q_2$.

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).

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