Question

Two particles having charges of 0.580 nC and 5.22 nC are separated by a distance of 1.80 m . At what point along the line connecting the two charges is the net electric field due to the two charges equal to zero?Where would the net electric field be zero if one of the charges were negative?

Answer 1

let
q1 = 0.58 nC
q2 = 5.22 nC
d = 1.80 m

let x is the distance from q1 where net elctric field is zero. Here the point must be between the two charges.

Apply,|E1| = |E2|

k*q1/x^2 = k*q2/(d - x)^2

(d-x)^2/x^2 = q2/q1

(d-x)/x = sqrt(q2/q1)

(1.8 - x)/x = sqrt(5.22/0.58)

1.8 - x = 3*x

1.8 = 4*x

==> x = 1.8/4

= 0.45 m from q1 <<<<<<<<-----------Answer

If one of the charges is negative.
let
q1 = -0.58 nC
q2 = 5.22 nC
d = 1.80 m

let x is the distance from q1 where net elctric field is zero. Here the point is not between the two charges.

Apply, |E1| = |E2|

k*q1/x^2 = k*q2/(d + x)^2

(d+x)^2/x^2 = q2/q1

(d+x)/x = sqrt(q2/q1)

(1.8 + x)/x = sqrt(5.22/0.58)

1.8 + x = 3*x

1.8 = 2*x

==> x = 1.8/2

= 0.90 m from q1 <<<<<<<<-----------Answer