Question

A 2.20 10-9 C charge has coordinates x = 0, y = −2.00; a 3.18 10-9 C charge has coordinates x = 3.00, y = 0; and a -4.60 10-9 C charge has coordinates x = 3.00, y = 4.00, where all distances are in cm. Determine magnitude and direction for the electric field at the origin and the instantaneous acceleration of a proton placed at the origin.

(a) Determine the magnitude and direction for the electric field at the origin (measure the angle counterclockwise from the positive x-axis). magnitude =

direction = °

(b) Determine the magnitude and direction for the instantaneous acceleration of a proton placed at the origin (measure the angle counterclockwise from the positive x-axis). magnitude =

direction = °

Answer 1

q1 = 2.20*10^-9 C (0,-2)
q2 = 3.18*10^-9 C (3, 0)
q3 = -4.6* 10^-9 C (3, 4),

Electric Field due to point charge is given by,
E = k*q1/r^2
E1 = (8.9*2.20)/(0.02^2) j^
E1 = 48950 j^

E2 = k*q2/r^2
E2 = - (8.9*3.18)/(0.03^2) i^
E2 = - 31446.6 i^

E3 = k*q3/r^2
E3 = (8.9*4.6)/(0.05^2) * 3/5 i^ + (8.9*4.6)/(0.05^2) * 4/5 j^
E3 = 9825.6 i^ + 13100.8 j^

Enet = E1 + E2 + E3
Enet = (- 31446.6 +9825.6) i^ + (48950 +13100.8 )j^
Enet = -21621 i^ + 62050.8 j^

Magnitude of the Electric Field, E = 65709.7 N/C
Direction of the Electric Field, = 109 Counterclockwise from +ve x-axis

(b)
We know,
F = q*E
F = 1.6 * 10^-19 * 65709.7 N
F = 1.05 * 10^-14 N

a = F/m
a = (1.05 * 10^-14)/(1.67*10^-27) m/s^2
a = 6.29 * 10^12 m/s^2

Magnitude of the instantaneous acceleration, a = 6.29 * 10^12 m/s^2
Direction of the instantaneous acceleration, = 109 Counterclockwise from +ve x-axis