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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 .600nC

i found out the answer to this to be .325m. by doing the following:

E due to q1 is equal in magnitude to E due to q2 but in opposite direction

As both q1 and q2 are positive, the field is zero at a point in between them

As distance from q1 is L , distance from q2 is (1.3 - L)

kq1/L^2 = kq2/(1.2 - L)^2

(1.3 - L) / L= sq rt (q2 / q1)

(1.3 - L) = L* sq rt (q2 / q1)

1.3 = L [1+sq rt (q2 / q1)]

L = 1.3 / [1+sq rt (q2 / q1)]

L = 1.3 / [1+sq rt (5.40 / 0.600)]

L = 1.3 / [1+sq rt ( 9.00 )]

L = 1.3 / [1+ 3.0]

L = 1.3 / [ 4.0]

L=0.325 meter from 0.550 nC

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

Enter answer as a distance from the charge initially equal 0.600 nC . This is where i get lost, i don't know what to do next. please Help!

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