# Electric Field and Distance From Particle

## Homework Statement

Two particles with positive charges q_1= 0.540 nC and q_2= 8.20 nC are separated by a distance of 1.30 m.

Along the line connecting the two charges, at what distance from the charge q_1 is the total electric field from the two charges zero?

## Homework Equations

E = k*q/r^2

d = [-b +/ sqrt(b^2 - 4ac)]/[2a]

## The Attempt at a Solution

I tried drawing a picture:

q_1 (0.540 nC) O____>____.____<____O q_2 (8.20 nC)

O = charged particle
. = field point
The arrows show the direction of the electric fields. (Correct????)

x+y = 1.30 m
y = 1.30 m – x

E = (k*q_1)/(x^2) – (k*q_2)/(y^2)
0 = (k*q_1)/(x^2) – (k*q_2)/(y^2)
0 = (k*q_1)/(x^2) – (k*q_2)/(1.30-x)^2

0 = (k*q_1)/(x^2) – (k*q_2)/(x^2-2.60x+1.69)

(k*q_2)/(x^2-2.60x+1.69) = (k*q_1)/(x^2)

Simplifying:

x^2*(k*q_2) – k*q_1(x^2-2.60x+1.69) = 0

x^2*(k*q_2) – (k*q_1*x^2) + (2.60*x*k*q_1) – 1.69*k*q_1 = 0

or

68.848x^2 + k*q_1*(2.60x –1.69) = 0

Using quadratic equation, x = 0.265 m or –0.449 m

Distance from q_1 = 0.265 m????????

Thanks.

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Gokul43201
Staff Emeritus
$$q_1/r_1^2 = q_2/r_2^2 \implies r_1/r_2 = \sqrt{q_1/q_2} = 0.257$$
$$\implies r_1+r_2 = 1.257r_2=1.3m \implies r_2=1.035m~,~~r_1=0.265m$$