Question

Two charges of value 2 microcoulomb and -50 microcoulomb are placed 80 cm apart calculate the distance of the point from the small the charge where the intensity is zero

Answer 1

20 cm distance from the smaller charge where intensity becomes zero.

let x distance from the smaller charge, electric field intensity becomes zero.

as it is given both charges are opposite in nature. so, point where electric field intensity becomes zero lies outside.

so, electric field at the point due to smaller charge + electric field at the point due to higher charge = 0

⇒k(2μC)/x² + [-k(50μC)/(80 + x)²] = 0

⇒2k/x² = 50k/(80 + x)²

⇒1/x² = 25/(80 + x)²

⇒5x = (80 + x)

⇒5x - x = 80

⇒x = 80/4 = 20cm

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Answer 2

$\huge\bold\purple{Answer:-}$

Let the distance between two point charges q and - 3q is r.

now electric field due to -3q at the location of q

is given by , E = k(-3q)/r² = (-3)kq/r² .....(1)

again, electric field due to q at the location of -3q is given by , E' = kq/r² ......(2)

from equations (1) and (2), it is clear that

E = -3E'

so, E' = -E/3 ,

hence, electric field due to q at the location of -3q is -E/3