Question

Two charges -20uC (2x10^-6) and +100uC are separated by a distance of 2 meters.

A) find the electric field midway between the two charges(state magnitude and direction) electric field is a vector quantity.

B) if you were to plop a +5uC charge right at this position what would be the electrical force on it?

Answer 1

here,

q1 = - 20 uC = - 20 * 10^-6 C

q2 = 100 uC = 100 * 10^-6 C

distance , r = 2 m

a)

the electric field , E = K * q1 /(r/2)^2 + K * q2 /(r/2)^2

E = 9 * 10^9 * 10^-6 /(2/2)^2 * ( 20 + 100)

E = 1.08 * 10^6 N/C

the direction of field is towards the negative charge (- 20 uC )

b)

q = 5 uC = 5 * 10^-6 C

the electric force on q , F = q * E

F = 5 * 10^-6 * 1.08 * 10^6 N

F = 5.4 N

the direction of force is towards the negative charge ( -20 uC)