Question

A + 7.9 nC point charge and a - 4.8 nC point charge are 4.15 cm apart. What is the electric field strength at the midpoint between the two charges?

Answer 1

In order to solve this, we need to first find the magnitudes of the E-field of each point charge

We can use the formula

E = kq/r^2 to find the E-field

E1 = (9x10^9)(7.9x10^-9)/(2.075 x10^-2)^2 = 165132.8 V/m

E2 = (9x10^9)(4.8x10^-9)/(2.075 x10^-2)^2 = 100333.9 V/m

Since one particle is negative and the other is positive, add the E-fields

Etotal = 165132.8 + 100333.9 = 265466.686 V/m