Question

The strength of an electric field at 5.0 cm from a point charge is 100.0 N/C. What is the magnitude of the source charge?
Please explain your answer.

Answer 1

Coulumbs law, force between 2 charges, q1, q2, separated by distance r:

F = k q1 q2 /r^2

the field strength E is measured as F/q, force per charge, so the field around a charge q, at distance r from charge, is:

E = k q /r^2

Answer 2

Thank you!

Answer 3

Can u tell me charge q=?

Answer 4

Can yu tell me k=?

Answer 5

E = k q /r^2
in given problem
E= 100.0 N/C
r=5.00 cm= 0.05 m
k=9x10^9
so equation becomes
100 = 9 x 10^9x q/ 0.05^2
we get
q= 2.7x10^-11 C

Answer 6

k = 4 = 9*10^9
q = 100*5*5/10000*9*10^9 = 2.7 * 10^(-11)