Question

2 charges as shown are placed 10 m apart. Where should the 3rd charge be placed from 'q' between them so that net force on this charge is 0. (With steps plz)
A) 10/3 m
B) 20/3 m
C) 10 m
D) none of these

Answer 1

Explanation:

hope that's useful and u understood mate

Answer 2

Explanation:

Force = k q₁q₂/r², where q₁ and q₂ are charges and r is  distance between them.

 Here, let the third should be placed at x distance form 4q C charge. Hence, 10 - x distance from other one.

As these charges are +, that must be -ve , to attract them, so that the net force become 0. Let that be Q

⇒ k 4q*Q/x^2 = k q*Q/(10 - x)^2

⇒ 4(10 - x)^2 = x^2

⇒ 400 + 4x^2 - 80x = x^2

⇒ 3x^2 - 80x + 400 = 0

⇒ 3x^2 - 60x - 20x + 400 = 0

⇒ (3x - 20)(x - 20) = 0

⇒ x = 20/3       or  x = 20

        It must not be outside, so it should be placed at 20/3 m   away from 4q charge.

Or, 10 - 20/3 = 10/3 m   away from q C.