Question

two charges 5*10-8 and _3*10-8C are located 16 cm apart .At what points on the line joining the two charges is the electric potential zero?

Answer 1

There are two charges,

Distance between the two charges, d = 16 cm = 0.16 m

Consider a point P on the line joining the two charges, as shown in the given figure.

 

r = Distance of point P from charge q1

Let the electric potential (V) at point P be zero.

Potential at point P is the sum of potentials caused by charges q1and q2 respectively.

Where, = Permittivity of free space

For V = 0, equation (i) reduces to

Therefore, the potential is zero at a distance of 10 cm from the positive charge between the charges.

Suppose point P is outside the system of two charges at a distance s from the negative charge, where potential is zero,

as shown in the following figure.

For this arrangement, potential is given by,

For V = 0, equation (ii) reduces to

Therefore, the potential is zero at a distance of 40 cm from the positive charge outside the system of charges.

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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