Two positive point charges 400 mu C and 200 mu C are placed

Two positive point charges +4.00 mu C and +2.00 mu C are placed at the opposite comers of a rectangle as shown in the figure. (k -=1/4 pi epsilon_0 = 8.99 times 10^9 N middot m^2/C^2) (a) What is the potential at point A (relative to infinity) due to these charges? (b) Two charges of - 2.00 mu C arc placed at points A and B. Calculate the net force on the +4.00mu C In the figure, a small spherical insulator of mass 6.00 times 10^-2 kg and charge +0.400 mu C is hung by a thin wire of negligible mass. A charge of -0.220 mu C is held 0.290 m away from the sphere and directly to the right of it, so the wire makes an angle theta with the vertical (k = 1/4 pi epsilon_0 = 8.99 times 10^9 N middot m^2/C^2)

Solution

(A) potential at point distance d from point charge q is given as

V = k q / d

at A:

V = [(9 x 10^9 x 2 x 10^-6)/ 0.400] + [(9 x 10^9 x 4 x 10^-6) / 0.8]

= 90000 volt .....Ans

(B) force on 4uC due to charge at A:

F1 = (9 x 10^9 x 4 x 10^-6 x 2 x 10^-6)/(0.8^2) (-i) = - 0.1125i N

due to charge at B:

F2 = (9 x 10^9 x 4 x 10^-6 x 2 x 10^-6)/(0.4^2) (-j) = - 0.450 j N

due to +2uC:

F3 = (9 x 10^9 x 4 x 10^-6 x 2 x 10^-6)/(0.4^2 + 0.8^2) (cos@i + sin@j)

and @ = tan^-1(0.4/0.8) = 26.6 deg

F3 = 0.09 ( cos26.6i + sin26.6) = 0.0805i + 0.0403 j N

Fnet = F1 + F2 + F3 = - 0.032i - 0.4097j N

magnitude of Fnet = sqrt(0.032^2 + 0.4097^2) = 0.411 N ........Ans