Question

Two fixed charges, -3.6?C and -5.0?C , are separated by a certain distance. If the charges are separated by 26cm , calculate the magnitude of the net electric field halfway between the charges.

Answer 1

The mid point between the charges is at 13 cm from each charge.

Let -3.6x10^-6 C be on the left of the mid point.

The electric field is the force experienced by a unit positive charge placed at the mid point.

Let F1 and F2 be the force experienced by the unit positive charge .
F2 > F1
F2 is towards the right and F1 is towards the left.

By Coulomb law,

F = k q1q2 / r^2
So the net force experienced by the unit positive charge placed at the mid point is ,

F2 - F1 = 9 x10^9 ( 5-3.6 ) x10^-6 / (13 x10^-2)^2 = 745.6 x10^3 N (towards the right)

Hence the electric field E at the mid point is

E = 745.6 x10^3 N/C

Answer 2

The mid point between the charges is at 30 cm from each charge.

Let -3.6x10^-6 C be on the left of the mid point.

The electric field is the force experienced by a unit positive charge placed at the mid point.

Let F1 and F2 be the force experienced by the unit positive charge .
F2 > F1
F2 is towards the right and F1 is towards the left.

By Coulomb law,

F = k q1q2 / r^2
So the net force experienced by the unit positive charge placed at the mid point is ,

F2 - F1 = 9 x10^9 ( 5 - 3.6 ) x10^-6 / (13 x10^-2)^2 = 745.5x10^3 N (towards the right)

Hence the electric field E at the mid point is

E = 774.5 x10^3 N/C ... (towards the right)