Online HW 1 Begin Date: 1/19/2018 4:35:00 PM -- Due Date: 1/26/2018 5:00:00 PM E
ID: 1661880 • Letter: O
Question
Online HW 1 Begin Date: 1/19/2018 4:35:00 PM -- Due Date: 1/26/2018 5:00:00 PM End Date: 1/26/2018 11:59:00 PM (1396) Problem 7: A positively charged particle (-15 nC is held fixed at the origin. A negatively charged particle O2=-16 nC of mass m = 6.5 g is located a distance d35 cm along the positive x-axis from the positively charged particle Q2 Otheexpertta.com × 33% Part (a) What is the magnitude of the electric force, F in newtons, that acts on the charge Q2? là 33% Part(b) What is the direction of the force on charge Q2? 33% Part (c) What is the magnitude of the acceleration, a in m/s2, experienced by charge Q2? Grade Su a= 090 100% Potential 7 8 9 coso) asin( atan()acotan)sinh( coshtanh) cotanh0 O Degrees O Radians sinO cotan tan() | acos Submissions Attempts remaining: 3 (4%per attempt) detailed view END BACKSPAC CLEAR Submit Hint I give up! Hints: 4% deduction per hint. Hints remaining: 3 Feedback: 5% deduction per feedback.Explanation / Answer
part a)
magnitude of electric force = k * Q1 * Q2/d^2
magnitude of electric force = 9 *10^9 * 15 *10^-9 * 16 *10^-9/0.35^2
magnitude of electric force = 1.76 *10^-5 N
b)
as the charges are of oposite , they will attract each other
the direction of force on Q2 is towards Q1
direction of Q2 is to the left
c)
acceleration of Q2 = force/mass
acceleration of Q2 = 1.76 *10^-5/(6.5 *10^-6)
acceleration of Q2 = 2.7 m/s^2
the acceleration of Q2 is 2.7 m/s^2
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.