Question

Three charged particles are at the corners of an equilateral triangle as shown in the figure below. (Let q = 2.00 mu C, and L = 0.600 m.) (a) Calculate the electric field at the position of charge q due to the 7.00-mu C and -4.00-mu C charges. late the magnitude of the field contribution from each charge you need to add these as vectors. kN/C i^Hat + ___________ kN/C J^Hat (b) Use your answer to part (a) to determine the force on charge q. ________ mN i^Hat + __________ mN j^Hat

Answer 1

a)lets first calculate the field due to 7 uC

E1x = k q/l^2 cos60

E1x = 9 x 10^9 x 7 x 10^-6 x cos60/0.6^2 = 8.75 x 10^4 N/C

E1y = k q/l^2 sin60

E1y = - 9 x 10^9 x 7 x 10^-6 x sin60/0.6^2 = -15.16 x 10^4 N/C

Field due to -4 is along x only

E2x = 9 x 10 x 4 x 10^-6 x cos0/0.6^2 = 10 x 10^4 N/C

So the total field will be:

E = [(10 - 8.75) x 10^4 ]i + 15.16 x 10^4 j = 12.5 kN/C i - 151.6 kN/C j

E = 12.5 kN/C i - 151.6 kN/C j

b)Force on charge is :

F = q E

F = 2 x 10^-6 [1.75 x 10^4 N/C i - 15.16x 10^4 N/C j]

F = 0.35 mN i - 0.303 mNi