Question

a particle at the origin of a cartesian coordinate system carries a charge of 3.89×10^-9 C.
1. What is the magnitude of the electric field at (4.50 mm, 0)?
2. what is the direction of the electric field at (4.50mm,0)?
3. what is the magnitude of the electric field st (0, 4.50)?

Answer 1

electric field is a vector quantity, which is given by:

E = kQ/R^2

Direction of E, due to positive charge (alpha) is away from charge and due to negative charge (electron) is towards the charge, So

Part 1.

Electric field at P (4.50 mm, 0) will be:

Q = 3.89*10^-9 C

R = distance between origin and P = 4.50 mm = 4.50*10^-3 m

E = 9*10^9*3.89*10^-9/(4.50*10^-3)^2

E = 1.73*10^6 N/C = Magnitude of electric field

Part 2.

Since charge is positive, So electric field at P will be away from the charge toward +ve x-axis (0 deg CCW from +ve x-axis)

Part 3.

Electric field at P (0, 4.50 mm) will be:

Q = 3.89*10^-9 C

R = distance between origin and P = 4.50 mm = 4.50*10^-3 m

E = 9*10^9*3.89*10^-9/(4.50*10^-3)^2

E = 1.73*10^6 N/C = Magnitude of electric field

Direction =

Since charge is positive, So electric field at P will be away from the charge toward +ve y-axis (90 deg CCW from +ve x-axis)

Please Upvote.