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

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)

the force represented by vector a is normal force. if the hockey puck is moving to the right, when it is hit with a significant force by the hockey stick the puck will move off in a different direction. correct on edge.