Question

Two point charges of equal magnitude (and opposite sign) are 7.5 cm apart. At the midpoint of the line connecting them, their combined electric field has a magnitude of 45 N/C. What is the magnitude of each charge?

Answer 1

Comment

The only reason you can do this is that the charges are the same. If they were not, the problem would not be possible.

Equation

The field equation is, in its simplest form,

E = kq/r^2

So each of the charges are pulling / pushing in the same direction. The equation becomes.

kq/r^2 - (-kq/r^2) = Field magnitude in N/C

Givens

• K = 9 * 10^9 N m^2 / c^2
• E = 45 N/C
• r = 7.5/2 = 3.75 cm * ( 1 m / 100 cm) = 0.0375 m
• Find Q

Solution

k*q/0.0375 ^2 - (-kq/0.0375^2) = 45 N/C Combine

2*k*q / 0.0375^2 = 45 N/C Divide by 2

kq /(0.0375^2) = 22.5 N/C Multiply by 0.0375^2

kq = 22.5 * 0.0375 ^2 Find d^2

kq = 22.5 * 0.001406 Combine

kq = 0.03164 N/C * m^2 Divide by k

q = 0.03164 N * m^2 /C / 9*10^9 N m^2 / c^2

q = 2.84760 * 10 ^8 C

I've left the cancellation of the units for you. Notice that only 1 C is left and it is in the numerator as it should be.