The Magnitude and the Direction of an Electric Field

Let the test charge +q be placed midway on line joining the positions of two charges.

Force between - 8.0 uC and test charges,F1 = (K 8.0*10^-6*q)/(0.02)^2 N
Direction of force(attraction) - towards - 8.0 uC charge.

Force between +6.0 uC and test charge, F2 = (k 6.0*10^-6*q)/(0.02)^2 N
Direction of force(repulsion) toward -8.0uC charge along the line joining two charges.

Net force F on the test charge q = F1 + F2
F = {K*10^-6*q (8 + 6)}/(0.02)^2 N
F/q ={k*10^-5}/0.0004 ...........................(F/q is force per unit charge= E)
E = {K*14*10^-6}/(4*10^-4) ........... (1)
= K*3.5*10^-2 N/C
= 9*10^9*3.5*10^-2
= 31.5*10^7 N/C
= 31.5*10^7 N/C
= 3.15*10^8 N/C
E is directed towards the - 8.0 uC charge along the line joining the two charges.

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