Help 1 Contact UsT Log INTRO TO PHYSICS (PHYS 207/2 Cutnell, Physics, 9e BACK Ch

Question

Help 1 Contact UsT Log INTRO TO PHYSICS (PHYS 207/2 Cutnell, Physics, 9e BACK Chagter 19, Problem 04 G0 Your answer is partialy correct. Try again. A partcle has a charge of +2.5 uC and moves from point A to point B, a distance of 0.17 m. The particle experiences a The difference between the particle's electric potential of the electric field that the particle experiences of 0.17 m. The partidle experiences a constant electric force, and its motion is along the line of action of the force b i energy at A and 8 is EPEA EPE - +8.4x10 J. () Find the magnitude of the electric force that acts on the particle. (b) Find the magnitude (e) Number 44-3 Units N (b) Number-1976.47 Units N/C Question Attempts: 1 of 3 used SAVE FOR LATER OR LATERSUBMIT ANSWER

Explanation / Answer

VA-VB= difference in EPE / q
= 8.4*10^-4 / (2.5*10^-6)
= 336 V

E = Difference in V/d
= 336 / 0.17
= 1976.47 N/C

E will be positive as it is directed from A to B

F= q*E
= (2.5*10^-6)*(1976.47)
= 4.94*10^-3 N

Answers:
4.94*10^-3 N
1976.47 N/C

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