Question

Two particles having charges q1 = 0.400 nC and q2 = 6.00 nC are separated by a distance of 1.60 m .

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

Answer 1

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:

0.4/r^2 = 6/(1.6-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:

(1.6-r)^2 = 15r^2
=>
1.6-r = 3.873r
=>
1.6 = 4.873r
=>
r = 1.6/4.873 = 0.413 m

so The electric fields are equal when you are 0.413 m away from the 0.4 nC charge and since the electric fields act in opposite directions (both are positively charged particles) the field will equal zero there.