Question

two point charges 20 microcoulomb and -4 microcoulomb are separated by a distance of 50 cm in air find the point on the line joining the surface where the potential is zero, calculate the electrostatic potential energy of the system​

Answer 1

q1=20uC

q2=-4uC

V1=V2

kq1/r=kq2/(0.5-r)

20/r=4/(0.5-r)

2.5-5r=r

6r=2.5

r=2.5/6m

Answer 2

Given :

Two point charges 20 mC and -4 mC .

Separation between charges ,  d = 50 cm = 0.5 m .

To Find :

The point on the line joining the surface where the potential is zero .

The electrostatic potential energy of the system​ .

Solution :

We know , potential due to charge q is given by :

$P=\dfrac{kq}{r}$

Let , x be the distance from charge 20 mC when potential is zero .

Therefore , distance of -4 mC is 50 - x .

Therefore ,

$P_{20}+P_{-4}=0\\\\\dfrac{k(20)}{x}+\dfrac{k(-4)}{50-x}=0\\\\20(50-x)-4x=0\\\\x=41.67\ cm$

Therefore , at the distance of 41.67 cm from 20 mC charge the potential is zero .

Now , potential energy is given by :

$U=\dfrac{9\times 10^9\times(20\times 10^{-6})\times (-4\times 10^{-6})}{0.5}\\\\U=-1.44\ J$

Learn More :

Electrostatics

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