Question

A uniform electric field exists everywhere in the x,y plane. The electric field has a magnitude of 3500 N/coil, and is directed in the positive x direction. A point charge of -9.0 x 10-9 coil is placed at the origin. Determine the magnitude of the net electric field at: (a) x

Answer 1

5525 N/C

Step-by-step explanation:

Magnitude of electric field ( E ) = 3500 N/c

Direction of electric field : positive X axis

point charge ( q ) = -9.0 * 10^-9

Calculate the Magnitude of the net electric field at (a) x = -0.20 m

Magnitude = 5525 N/C

Electric field due to q = ( 9 * 10^9 * 9 * 10^-9 ) / ( -0.2 )^2

= 81 / 0.04 = 2025 N/c

Therefore the magnitude of the net electric field

= 2025 + 3500

= 5525 N/C