Question

Two particles having charges of 0.480nC and 7.68nC are separated by a distance of 1.70m .

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

B: Where would the net electric field be zero if one of the charges were negative? Enter your answer as a distance from the charge initially equal 0.480nC

Answer 1

Here is what I solved before, please modify the figures as per your question.

The electric field generated by a point charge is

E = (Q/r^2)(1/((4)(pi)(e)))

Where e is the permittivity of free space.

You can set up the two fields and make them equal to each other:

(Q1/r^2)(1/((4)(pi)(e))) = (Q2/(1.2-r)^2)(1/((4)(pi)(e)))

We need to solve for r, where r is the distance from Q1.

the (1/((4)(pi)(e))) cancels on both sides to get:

Q1/r^2 = Q2/(1.2-r)^2 plug in values for Q1 and Q2:

.5/r^2 = 8/(1.2-r)^2 (notice the units for charge doesn't matter since they will cancel out as long as they are in the same units) solve for r:

(1.2-r)^2 = 16r^2
=>
1.2-r = 4r
=>
1.2 = 5r
=>
r = 1.2/5 = .24m

so The electric fields are equal when you are .24m away from the .5nC charge and since the electric fields act in opposite directions (both are positively charged particles) the field will equal zero there.