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

`0.002 N/C`

Step-by-step 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 = (-ma)/(q)`

Solving for E, we have:

`E = (-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.