Question

two charges 5×10^(-8) C and -3×10^(-8) are located 16cm apart. at what points on the line joining the two charges is the electric potential zero ? take the potential at infinity to be zero.​

Answer 1

Step-by-step explanation:

Answer:

$\sf \: {\blue{ \underline{ \underline{ANSWER }}}}$

Given :

$\sf \: q_1 \: = \: 5 \times {10}^ {- 8}$

$\sf \: q_2 = - 3 \times {10}^{ - 8}$

Distance(d) = 16cm = 0.16m

To Find :

Position where the electric potential is 0.

Solution :

When, Electric Potential is zero.

$\implies \: \sf \: 0 = \: \frac{kq_1}{x} + \frac{kq_2}{(d - x)}$

$\implies \: \sf \: \frac{kq_1}{x} = - \frac{kq_2}{(d - x)}$

$\implies \: \sf \: \frac{ \cancel{k}q_1}{x} = - \frac{ \cancel{k}q_2}{(d - x)}$

$\implies \: \sf \: \frac{q_1}{x} = - \frac{q_2}{(d - x)}$

On putting the values,

$\implies \: \sf \: \frac{5 \times {10}^{ - 8} }{x} = \frac{ - ( - 3 \times {10}^{- 8})}{0.16 \: - \: x}$

$\implies \: \sf \: \frac{5 \times \cancel{{10}^{- 8}}}{x} = \frac{ 3 \times \cancel{{10}^{- 8}}}{0.16 \: - \: x}$

On Cross Multiplication, we get

$\implies \sf \: \: 3x = \: 0.80 \: - \: 5x$

$\implies \sf \: \: 8x = \: 0.80$

$\implies \: \sf x \: = 0.10 \:$

$\boxed{ \sf \: \therefore \: x =0 .10 \: m \: or \: 10 \: cm}$

Thus, the Electric Potential is zero at 10cm at right side of charge q(1) or (16 - 10) = 6cm at left side of charge q(2).