Two particles having charges of 0.570nC and 20.5nC are separated by a distance o

Question

Two particles having charges of 0.570nC and 20.5nC are separated by a distance of 2 m.

A. At what point along the line connecting the two charges is the net electric field due to the two charges equal to zero?

The electric field is zero at a point =________ m from 0.570nC.

B. Where would the net electric field be zero if one of the charges were negative?

(eneter answer as a distance from the charge initially equal 0.570nC)

_______m

For part A my answer is 0.286 but I can't figure out part B!! Please Help!

Explanation / Answer

Let me give you a detailed answer so as to give you a better understanding

1) Electric field is given by E = Kq/r^2

so Let the electric field is zero at x metres from 0.570 nC, so its distance from 20.5 nC will be (2-x) metres

so E due to 0.570 nC at this point = K x 0.570 E-9 / x^2

and E due to 20.5 nC at this point = K x 20.5 E-9 / (2-x)^2

Please note that both the fields will be in opposite direction

so for net electric field to be zero at this point,

E due to 0.570 nC at this point = E due to 20.5 nC at this point

thus K x 0.570 E-9 / x^2 = = K x 20.5 E-9 / (2-x)^2

so (2-x)^2 / x^2 = 20.5 / 0.570

this gives x = 0.2858 m

2) Let the point is at x metres from 0.57 nC charge.. this time the distance of this point from 20.5nC will be (2+x) as the point cannot be between them

so so E due to 0.570 nC at this point = K x 0.570 E-9 / x^2

and E due to 20.5 nC at this point = K x 20.5 E-9 / (2+x)^2

so for net electric field to be zero at this point,

E due to 0.570 nC at this point = E due to 20.5 nC at this point

thus K x 0.570 E-9 / x^2 = = K x 20.5 E-9 / (2+x)^2

so (2+x)^2 / x^2 = 20.5 / 0.570

or (2+x) = 5.997 x

this gives x = 0.4 m

Please note that this point is along the line joining the two charges but not between them, it is closer to 0.57nC charge at 0.4 metres and at 2.4 metres from 20.5nC charge

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