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)

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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