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What third charge should be placed at x=+26cm so that the total electric field at x=+13.0cm is zero?

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Complete text of the problem:

"Two point charges lie on the x axis. A charge of + 2.20 pC is at the origin, and a charge of − 4.60 pC is atx=−13.0cm.

What third charge should be placed at x=+26cm so that the total electric field at x=+13.0cm is zero?

Express your answer to three significant figures and include appropriate units"

Let's call A the point at x=+13 cm where the total electric field is zero.

First of all, we can calculate the total field generated by the two charges
q_1=2.20~pC=2.2\cdot10^(-9)~C and
q_2=-4.6~pC=4.6\cdot10^(-9)~C. The two charges have a distance from point A of
r_1=13~cm=0.13~m and
r_2=26~cm=0.26~m, respectively. The field generated by the positive charge heads towards right in point A, while the one generated by the negative charge heads towards left, so we should consider a sign - on this field. Therefore, the total field generated by charge 1 and 2 in A is


E_(12)=k_e (q_1)/(r_1^2)-k_e (q_2)/(r_2^2)=1170.3~V/m - 611.7~V/m=558.6~V/m

In order to have total net field of zero in point A, the field generated by charge 3 should be equal to this value, and should point towards the left, so it must be a positive charge. So, we have:


E_3 = E_(12)=558.6~V/m=k_e (q_3)/(r_3^2)
where
r_3=0.13~m is the distance of the charge 3 from point A. So we find


q_3=E_(12) (r_3^2)/(k_e)=1.05\cdot 10^-9~C=1.05~pC


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