Answer to Question #96744 in Electricity and Magnetism for Abru

Question

(electrostatics).
Figure below shows a charge q1 (= +1.0 x 10-6 C) 10.0 cm from a charge q2 (= -1.0 x 10-6 C).
(a) Find the electric field strength (magnitude and direction) at point P if the distance between charge q2 and point P is 2.5 cm.
(b)Find the electric force (magnitude and direction) on a test charge of 10.0 C at P.
(Coulomb constant: k = (1/40) = 8.9875109 Nm2/ C2)

Answer 1

(a) The field between these two charges:

$$E=E_1-E_2=k\bigg(\frac{q_1}{r_1^2}-\frac{q_2}{r_2^2}\bigg)=\\ \space\\ =9\cdot10^9\bigg(\frac{1\cdot10^{-6}}{0.075^2}-\frac{-1\cdot10^{-6}}{0.025^2}\bigg)=16\space000\space000\text{ V/m}.$$

(b) The force will be

$$F=Eq=16\space000\space000\cdot10=160\space000\space000\text{ N}.$$