Ask Question
20 December, 18:29

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?

+2
Answers (1)
  1. 20 December, 18:46
    0
    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)
Know the Answer?
Not Sure About the Answer?
Find an answer to your 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 ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers