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 `\frac {q₁}{x}` -----------(2)


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

Vq₂ = - k `\frac {q₂}{r - x}` ----------------(3)

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

k `\frac {q₁}{x}` = - k `\frac {q₂}{r - x}`

`\frac {q₁}{x}` = - `\frac {q₂}{r - x}`

`\frac {q₂}{q₁}` = -`\frac {r - x}{x}`

`\frac {q₂}{q₁}` = -(`\frac {r}{x}` - `\frac {x}{x}`)

`\frac {q₂}{q₁}` = -`\frac {r}{x}` + 1

-`\frac {r}{x}` + 1 = `\frac {q₂}{q₁}`

-`\frac {r}{x}`  = `\frac {q₂}{q₁}` - 1

-`\frac {r}{x}`  = `\frac {q₂ - q₁}{q₁}` 

taking -1 common from RHS

-`\frac {r}{x}`  = -1 (`\frac {-q₂ + q₁}{q₁}` )

by cross-multiplication we get

x = `\frac {q₁r}{q₁ - q₂}` 

by putting values

x = `\frac {8 ｘ 10⁻⁶ C ｘ0.1 m}{8 ｘ 10⁻⁶ C  - (- 4 ｘ 10⁻⁶ C) }` 

x = `\frac {0.8 ｘ 10⁻⁶ C m}{12 ｘ 10⁻⁶ C }` 

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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