The magnitude of the electric field at a certain location is

The magnitude of the electric field at a certain location is 500 N/C and the field is directed east to west, (a) Find the magnitude and direction of the force acting on a proton placed at this point, (b) Repeat for an electron located at the point. Find the magnitude and direction of the electric field at a distance of 10.0 cm from an electron.

Solution

part A

force on proton(positive charge) is in the direction of electric field

F = Q*E

for proton Q = 1.6x10^-19 C

E = 500N/C

F = 1.6x10^-19 * 500

F = 8x10^-17 N towards west

Part B

for electron force magnitude of force is same but direction is opposite as electron has negative charge

F = 8x10^-17 N towards east

Part C

electric field at 10cm

E = K*Q/d^2

K = 9*10^9

Q = 1.6*10^-19C

d = 10cm = 0.10m

E = 9*10^9 * 1.6*10^-19 / 0.10*0.10

E = 1.44 x 10^-7 N/C