Question

Two point charges of equal magnitude are 6.1 cm apart. At the midpoint of the line connecting them, their combined electric field has a magnitude of 37 N/C. Find the magnitude of the charges.

Answer 1

E=kq/r^2

Charges have to be of opposite sign

37=2*k*q/r^2

=> q=37/2/9/10^9*.061^2=7.648*10^-12 C

Answer 2

Since at the mid way the force exists the charges, equal in magnitude, have to be oppositely charged. If q be the magnitude then the force will be twice that exerted by any one of them at the mid point. ie 2* q * 9*10^9 / (3.25*10^--2) ^2 = 75 (given in the problem) Therefore q = 75*(3.25*10^--2) ^2 / 2*9*10^9 q = 4.4*10^--12 C Or q = 4.4 pC.

Answer 3

Since at the mid way the force exists the charges, equal in magnitude, have to be oppositely charged.

If q be the magnitude then the force will be twice that exerted by any one of them at the mid point.

ie 2* q * 9*10^9 / (3.05*10^--2) ^2 = 37(given in the problem)

Therefore q = 37*(3.05*10^--2) ^2 / 2*9*10^9

q = 1.91*10^-12 C

Or q = 1.91pC.

Answer 4

Since electric field exists at the mid point, charges have equal magnitude, but opposite charges. So,

E = kq/r