Question

What is the magnitude and direction of the force exerted on a 3.50 µC charge by a 250 N/C electric field that points due east? (answer in ×10^{-4} N)

Answer 1

Answer:

8.75×10⁻⁴ N. due east

Explanation:

Electric Field: This is defined as the force per unit charge which it exerts at that point. Its direction is the force exerted on a positive charge. The S.I unit of electric field is N/C.

The expression for electric field is given as,

E = F/q

where E = Electric field, F = Force, q = charge.

Making F the subject of the equation,

F = E×q.................... Equation 1

Given: E = 250 N/C, q = 3.5 µC = 3.5×10⁻⁶ C.

Substitute into equation 1

F = 250×3.5×10⁻⁶

F = 875×10⁻⁶

F = 8.75×10⁻⁴ N. due east

Hence the force is 8.75×10⁻⁴ N. due east

Answer 2

The magnitude and direction of the electric force on the charge is $8.75 \times 10^{-4} \ N$ upwards.

The given parameters;

• charge, q = 3.5 µC  
• electric field, E = 250 N/C

The magnitude of the electric force on the charge is calculated as follows;

$F = Eq\\\\F = (250) \times (3.5 \times 10^{-6})\\\\F = 8.75 \times 10^{-4} \ N$

Thus, the magnitude and direction of the electric force on the charge is $8.75 \times 10^{-4} \ N$ upwards.

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