Question

Two charges are placed on the x axis. One of the charges (q1 = +6.88C) is at x1 = +3.00 cm and the other (q2 = -25.3C) is at x2 = +9.00 cm. Find the net electric field (magnitude and direction given as a plus or minus sign) at (a) x = 0 cm and (b) x = +6.00 cm.

Answer 1

Are you sure the charges are in C ( They should be in uC) . Anyway , I will solve with what is given in the question. If you want you can solve with uC using same method.

k = 9e9 ( constant)

E = kq / r²

for charge at 3 m = 0.03 m

E = - kq₁ / r²

similarly for charge at x = 9 cm = 0.09 m

E (total) = - k ( 7644.44 - 3123.45)

E (total) = - 4.0688e13 N/C ( in negative x - direction)

b) For x = 6 cm

E (total) = k ( 6.88 / 0.03² - (-23.5 / 0.03²))

E (total) = k ( 7644.44 - (-26111.1)

E (total) = 3.037e14 N/C ( in positive x - direction )