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## Homework Statement

A negative point charge q_1 = -4.00 nC is on the x-axis at x = 0.60 m. A second point charge q_2 is on the x-axis at x = -1.20 m.

a. What must the sign and magnitude of q_2 be for the net electric field at the origin to be 50.0 N/C in the +x-direction?

b. What must the sign and magnitude of q_2 be for the net electric field at the origin to be 50.0 N/C in the -x-direction?

## Homework Equations

E = k*q/r^2

## The Attempt at a Solution

a. The charge must be negative for subtraction to occur.

(-q2) O__<____. _>__O (-q1)

-----E2 = (-)------E1= (+)

negative diagram

. = field point (0,0)

O = particle

Arrow indicates field direction based on positive test charge

E1 = (8.988*10^9)*(-4.00*10^-9 C)/(.60 m)^2 = 99.8667 N/C (+)

99.8667 N/C + E2 = 50 N/C

E2 = -49.8667 C

q_2 = r^2*(49.86667)/(8.988*10^9) = -7.9893*10^-9 C ?????

b. E1 = (8.988*10^9)*(-4.00*10^-9 C)/(.60 m)^2 = 99.8667 N/C (+)

E2 = -50 N/C – 99.8667 = -149.8667 C

q_2 = [(r^2)149.8667 N/C]/k = [(r^2)149.8667 N/C]/k = (1.2^2)*149.8667/(8.988*10^9) = -2.40*10^-8 C ??????

Thanks.