Question

Find the electric field 22.0 cm from the center of the charge distribution.

Answer 1

Coulomb's law:\[F= \frac{Qq }{ 4\pi \epsilon _{0}r^2 }\]

Answer 2

Q is the charge at the center q is the size of the second charge (in this case it is 0 since there is no second charge)

r is the distance from Q and \[\epsilon _{0}= 8.854 187 817... * 10^{-12}\]

Answer 3

\[\pi=3.14159...\]

Answer 4

the question is asking about the strength of the e field at 22 cm right?