Question

What is the magnitude and direction of an electric field that exerts a 2.00×10⁵ N upward force on a –1.75μC charge?

a) ( 1.14 X 10⁴ N/C ) upward
b) ( 8.57 X 10³ N/C ) downward
c) ( 1.14 X 10⁴ N/C ) downward
d) ( 8.57 X 10³ N/C ) upward

Answer 1

The magnitude of the electric field is 1.14 × 10^4 N/C and the direction is downward, as the electric field is in the opposite direction to the force on a negative charge.

Step-by-step explanation:

To find the magnitude and direction of the electric field (E) that exerts a given force on a charge, we can use the formula:

E = F / q

where F is the force exerted on the charge, and q is the value of the charge. Substituting the given values into the equation:

E = 2.00 × 10-5 N / (-1.75 × 10-6 C) = -1.14 × 104 N/C.

The negative sign indicates the direction of the field is opposite to the direction of the force. Since the force is upward, the direction of the electric field is downward. Therefore, we can conclude that the magnitude of the electric field is 1.14 × 104 N/C and the direction is downward.