Question

A small object with a 5.0-mC charge is accelerating horizontally on a friction-free surface at 0.0050 m/s2 due only to an electric field. If the object has a mass of 2.0 g, what is the magnitude of the electric field

Answer 1

Answer:

$0.002 N/C$

Explanation:

Parameters given:

Charge of object, q = 5 mC = $5 * 10^{-3} C$

Acceleration of object, a = $0.005 m/s^2$

Mass of object, m = 2.0 g

The Electric field exerts a particular force on the object, causing it to accelerate (Electrostatic force).

We know that Electrostatic force, F, is given in terms of Electric field, E, as:

F = qE

This means that the object exerts a force of -qE on the Electric force (Action with equal and opposite reaction).

The object also has a force, F, due to its acceleration a. This force is the product of its mass and acceleration. Mathematically:

F = ma

Equating the two forces of the object, we get:

-qE = ma

=> $E = \frac{-ma}{q}$

Solving for E, we have:

$E = \frac{-2 * 10^{-3} * 0.005}{5 * 10^{-3}} \\\\\\E = -0.002 N/C$

The magnitude will be:

$|E| = |-0.002| N/C = 0.002 N/C$

The electric field has a magnitude of 0.002 N/C.