Question

A + 18 nC charge is located at the origin.

What is the strength of the electric field at the position (x,y)=(−5.0cm,5.0cm)?

What is the strength of the electric field at the position (x,y)=(5.0cm,0cm)?

What is the strength of the electric field at the position (x,y)=(−5.0cm,−5.0cm)?

Answer 1

Case 1 :

x = - 5 cm = - 0.05 m

y = 5 cm = 0.05 m

r = Distance of location from the charge at the origin = sqrt(x² + y²) = sqrt((- 0.05)² + (0.05)²) = sqrt(2) (0.05) m

q = magnitude of the charge at the origin = 18 x 10^-9 C

Magnitude of electric field at the location is given as

E = k q/r²

E = (9 x 10⁹) (18 x 10^-9)/(sqrt(2) (0.05))²

E = 32400 N/C

Case 2 :

x = 5 cm = 0.05 m

y = 0 cm = 0 m

r = Distance of location from the charge at the origin = sqrt(x² + y²) = sqrt((0.05)² + (0)²) = (0.05) m

q = magnitude of the charge at the origin = 18 x 10^-9 C

Magnitude of electric field at the location is given as

E = k q/r²

E = (9 x 10⁹) (18 x 10^-9)/((0.05))²

E = 62800 N/C

Case 3:

x = - 5 cm = - 0.05 m

y = -5 cm = -0.05 m

r = Distance of location from the charge at the origin = sqrt(x² + y²) = sqrt((- 0.05)² + (- 0.05)²) = sqrt(2) (0.05) m

q = magnitude of the charge at the origin = 18 x 10^-9 C

Magnitude of electric field at the location is given as

E = k q/r²

E = (9 x 10⁹) (18 x 10^-9)/(sqrt(2) (0.05))²

E = 32400 N/C