2.1)  Two charges5 \times 10^{-8}C  and -3 \times 10^{-8}C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

Answers (1)
P Pankaj Sanodiya

Given, two charge particles

q_1= 5*10^{-8}C

q_2= -3*10^{-8}C

 The separation between two charged particle d=16cm=0.16m

Now, let's assume the point P between two charged particles where the electric potential is zero is x meter away from q_1 and ( 9-x) meter away from  q_2

So,

The potential at point P :

V_p=\frac{kq_1}{x}+\frac{kq_2}{0.16-x}=0

V_p=\frac{k5*10^{-8}}{x}+\frac{k(-3*10^{-8})}{0.16-x}=0

\frac{k5*10^{-8}}{x}=-\frac{k(-3*10^{-8})}{0.16-x}

5(0.16-x)=3x

x=0.1m=10cm

Hence the point between two charged particles where the electric potential is zero lies 10cm away from q_1 and 6 cm away from q_2

 

Now, Let's assume a point  Q which is outside the line segment joining two charges and having zero electric potential .let the point Q lie r meter away from q_2 and  (0.16+r) meter away from q_1

So electric potential at point Q = 0

\frac{kq_1}{0.16+r}+\frac{kq_2}{r}=0

\frac{k5*10^{-8}}{0.16+r}+\frac{k(-3*10^{-8})}{r}=0

5r=3(0.16+r)

r=0.24m=24cm

Hence the Second point where the electric potential is zero is 24cm away from q_2 and 40cm away from q_1

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