Question

Two positively charged particles each having charge 20microCoulomb, are kept at a distance

of 2.0 m from each other. Determine the force on each charge and the electric field

due to each charge. Show the force and electric field vectors on appropriate

diagrams. What is the resultant force at a point midway from the two charges along

the straight line joining them?​

Answer 1

Answer:

The electric field due to each charge can be found from the Coulomb's Law:

F_{e}=k\dfrac{Q_1Q_2}{r^2},F

e

=k

r

2

Q

1

Q

2

,

F_e=9\cdot10^9\ \dfrac{Nm^2}{C^2}\dfrac{(20\cdot10^{-6}\ C)^2}{(2.0\ m)^2}=0.9\ N.F

e

=9⋅10

9

C

2

Nm

2

(2.0 m)

2

(20⋅10

−6

C)

2

=0.9 N.

Since the two charges are both positively charged they will repel each other, therefore, the direction of the electric force is repulsive (also, the sign plus indicates that the direction is repulsive). Also, F_{e12}=F_{e21}=F_eF

e12

=F

e21

=F

e

.