Question

the acceleration of particle of charge of 0.5 C and 500 mg mass kept in an electric field of intensity 200 N/C is​

Answer 1

Answer:

The acceleration of charge particle.

Explanation:

Given:

Charge, q = 0.5C

Mass, m = 500 kg (mass in kg not in mg as shown in the question)

Electric field, E = 200 N/C

Using:

Electric field , it is defined as force per unit charge,

E =$\frac{F}{q}$

According to the Newton's second law we have,

F = ma

Putting this in equation (1) we get;

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

a = $\frac{Eq}{m}$

Calculations:

$a =\frac{100}{500} \\a=0.2$ m/$s^{2}$

Result:

The acceleration is equal to 0.2 m/s²