(2000) Problem 4: A charge of q = 3.8 x 10-16 C is placed in a constant electric
ID: 1652909 • Letter: #
Question
(2000) Problem 4: A charge of q = 3.8 x 10-16 C is placed in a constant electric field of E-2.1 N C. The charge is located at a distance r= 15 cm from the origin at an angle 15° above the x-axis. 9 Otheexpertta.com - 33% Part (a) Write an expression for the force, Fx, that the charge experiences in the x direction, in terms of the charge q and magnitude of the electric field E -a 33% Part (b) What is the force in the y-direction, Fy in Newtons? 33% Part (c) What is the magnitude of the force vector, F in N?Explanation / Answer
a)
a positive charge experience force in the direction of electric field. Since the electric field is in X-direction , hence the force on the charge by the electric field will also be in X-direction
so Fx = q E
Fx = (3.8 x 10-16) (2.1)
Fx = 7.98 x 10-16 N
b)
Fy = 0
c)
magnitude of force is given as
|F| = sqrt(Fx2 + Fy2) = sqrt((7.98 x 10-16)2 + (0)2) = 7.98 x 10-16 N
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.