Answer:
[tex]0.002 N/C[/tex]
Explanation:
Parameters given:
Charge of object, q = 5 mC = [tex]5 * 10^{-3} C[/tex]
Acceleration of object, a = [tex]0.005 m/s^2[/tex]
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
=> [tex]E = \frac{-ma}{q}[/tex]
Solving for E, we have:
[tex]E = \frac{-2 * 10^{-3} * 0.005}{5 * 10^{-3}} \\\\\\E = -0.002 N/C[/tex]
The magnitude will be:
[tex]|E| = |-0.002| N/C = 0.002 N/C[/tex]
The electric field has a magnitude of 0.002 N/C.
