Question

two charges e and 3e are placed at a distance r. The distance of the point where the electric field intensity will be zero is
$a)r \div (1 + \sqrt{3} )from 3e charge$
$b)r \div (1 + \sqrt{3} )from e charge$
$c)r \div (1 - \sqrt{3} )from 3e charge$
plz give answer with elaborate

Answer 1

3e…………….…(•)…………………………e
|<– – x – – –>|<– – – (r - x) – –>|

Where
(•) = The point where electric field is 0

Electric field due to 3e = Eelectric field due to e
k(3e) / (x)² = k(e) / (r - x)²
3/x² = 1/(r - x)²
Square root on both the sides
√3/x = 1 / (r - x)
x = √3(r - x)
x = r√3 - x√3
x + x√3 = r√3
x(1 + √3) = r√3
x = r√3 / (1 + √3) <= This is distance from 3e charge

Distance of (•) from e charge is
=> r - x
=> r - r√3 / (1 + √3)
=> r [1 - (√3 / (1 + √3))]
=> r [(1 + √3 - √3)/(1 + √3)]
=> r / (1 + √3)

∴ b) r ÷ (1 + √3) from e charge… is the correct option

Answer 2

Answer:

3e…………….…(•)…………………………e

|<– – x – – –>|<– – – (r - x) – –>|

Where

(•) = The point where electric field is 0

Electric field due to 3e = Electric field due to e

k(3e) / (x)² = k(e) / (r - x)²

3/x² = 1/(r - x)²

Square root on both the sides

√3/x = 1 / (r - x)

x = √3(r - x)

x = r√3 - x√3

x + x√3 = r√3

x(1 + √3) = r√3

x = r√3 / (1 + √3) <= This is distance from 3e charge

Distance of (•) from e charge is

=> r - x

=> r - r√3 / (1 + √3)

=> r [1 - (√3 / (1 + √3))]

=> r [(1 + √3 - √3)/(1 + √3)]

=> r / (1 + √3)

∴ b) r ÷ (1 + √3) from e charge… is the correct option