Two point charges of 8 µC and -4 µC are separated by a distance of 10 cm in air. At what point on the line joining the two charges is the electric potential zero? (Answer 6.6 cm from 8 µC and, 3.3 cm from -4 µC charge)  



Given:

Let 

1st point charge = q₁ = 8 µC = 8 ｘ 10⁻⁶ C 

2nd point charge = q₂ = -4 µC = -4 ｘ 10⁻⁶ C 

Separation distance = r = 10 cm = 0.1 m



To Find:

The distance at which E.P. is zero = d = ?

Solution:


The diagram can be drawn according to the given conditions are as below



Let at the point P the E.P. is considered to be zero. Let the distance from point P  to q₁ is x, then Electric Potential be,

Vq₁ = Vq₂ ----------------(1)

Vq₁ = k q₁x -----------(2)


Let the distance from point P  to q₂ is r - x, then Electric Potential be, 

Vq₂ = - k q₂r-x ----------------(3)

by putting Eqn (2) and (3) in Eqn (1) we get

k q₁x = - k q₂r-x

q₁x = - q₂r-x

q₂q₁ = -r-xx

q₂q₁ = -(rx - xx)

q₂q₁ = -rx + 1

-rx + 1 = q₂q₁

-rx  = q₂q₁ - 1

-rx  = q₂-q₁q₁ 

taking -1 common from RHS

-rx  = -1 (-q₂+q₁q₁ )

by cross-multiplication we get

x = q₁rq₁-q₂ 

by putting values

x = 8ｘ10⁻⁶Cｘ0.1m8ｘ10⁻⁶C  -(-4ｘ10⁻⁶C) 

x = 0.8ｘ10⁻⁶Cm12ｘ10⁻⁶C  

x = 0.06667 m

0r

x = 6.6667 cm from 8 µC ---------Ans.1

r - x = 10 cm - 6.6667 cm = 3.3 cm from -4 µC ------Ans.2



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