Question

A proton is released in a uniform electric field, and it experiences an electric force of 2.36×10−14 N toward the south. (a)What is the magnitude of the electric field? (b) What is the direction of the electric field?

Answer 1

Answer:

a) E = 1.47 × 10^5 N/C

b) south

Explanation:

The magnitude of an electric field can be defined mathematically as;

E = F/q ........1

Where,

E = magnitude of the electric field

F = electric force

q = charge on the proton

Given;

F = 2.36 × 10^-14 N

Note that charge on a proton is known as Qp = 1.602 × 10^-19 C

q = 1.602 × 10^-19 C

Substituting into equation 1, we have;

E = 2.36 × 10^-14 N/1.602 × 10^-19 C

E = 1.47 × 10^5 N/C

b) The direction of the electric field;

From equation 1

E = F/q ........1

since both electric field and electric force are vector quantity and q is a positive charge (constant), then both the electric field and electric force would be parallel to each other. Therefore the electric field is directed to the south also.

(When a vector is multiplied by a positive constant the direction remains the same)