Question

Two charges – 3 x 10-8C and 5 x 10–8C are located 0.16 m apart. At what point on the line joining
the two charges is the electric potential zero? Take the potential at infinity to be zero,​

Answer 1

Answer:

Let p be the point of zero potential be p at distance r from the charge 1.

d = 0.16 m = 16 cm

Electric potential,

$\large\rm { v = q_{1} / 4πϵ_{0}r + q_{2} / 4πϵ_{0} (d-r) }$

$\large\rm { for \ v = 0, }$

$\large\rm { \frac {q_{1}}{r} = - \frac {q_{2}}{d-r}}$

$\large\rm { ⇒ \frac{ 5 \times 10 ^ { -8 } }{ r } = \frac{ -3 \times 10 ^ { 8 } }{ 0.16-r } }$

$\large\boxed{\rm { r = 10 \ cm}}$

If, point p is outside,

$\large\rm { v = q_{1} / 4πϵ_{0}r + q_{2} / 4πϵ_{0} \ (r-d)}$

So here,

$\large\boxed{\rm { r = 40 cm}}$

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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