Question

What is the magnitude and direction of an electric field that exerts a 2.00×10−5N upward force on a –1.75μC charge?

Answer 1

Answer:

E = - 11.4 N/C, Directed Downward

Explanation:

Force = 2.00×10⁻⁵N , charge Q = –1.75μC = -1.75 ×10⁻⁶ C

To Find: Electric Field E = ? and Direction=?

Sol

we Have

E = F / Q

E = 2.00×10⁻⁵N /  -1.75 ×10⁻⁶ C

E = - 11.428 N/C (-ve sign shows that it is Negative Electric Field around a negative charge and will attract the Positive Field)

As given an upward force on negative charge so direction of the Electric Filed will be Downward (as Electric field is opposite to direction of force on negative charge)