Question

two charge 5 into 10 power minus 8 column and minus 3 into 10 power minus date column are located 16 cm apart at what point on line joining the two charges is electric potential zero take a potential at Infinity to zero

Answer 1

Let q1 = 5 × 10–8 C and q2 = –3 × 10–8 C

Use the formula , Potential V = q/4πεor

Suppose Potential is zero at a distance x cm from q1 hence from charge q2 it is (16-x) cm

⇒ V = o

⇒ [q1/4πεox × 10–2] + [q2/4πεo(16-x)× 10–2] = 0

On solving we get , x = 0.1 m = 10 cm

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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