PLEASE I ONLY NEED HELP WITH PART 2 AS YOU CAN SEE I HAVE THE ANSWERS IN FOR THE

Question

PLEASE I ONLY NEED HELP WITH PART 2 AS YOU CAN SEE I HAVE THE ANSWERS IN FOR THE FORST PART. PLEASE SHOW ME HOW TO DO THIS STEP BY STEP AND WITH CORRECT ANSWERS I WILL RATE 5 STAR

PRACTICE IT Use the worked example above to help you solve this problem. Charge q1 = 6.80 ?C is at the origin, and charge q2 = -5.20 ?C is on the x-axis, 0.300 m from the origin (see figure). (a) Find the magnitude and direction of the electric field at point P, which has coordinates (0, 0.400) m.

magnitude: 254950.9757N/C Correct: Your answer is correct.

direction: 64.4? Correct: Your answer is correct.

(b) Find the force on a charge of 2.00 x 10^-8 C placed at P.

Correct: 0.005099 N Your answer is correct

*****PART 2******

(a) Place a charge of -9.60 ?C at point P and find the magnitude and direction of the electric field at the location of q2 due to q1 = 6.80 ?C and the charge at P. magnitude N/C direction ? (b) Find the magnitude and direction of the force on q2. magnitude N direction ?

Explanation / Answer

force due to q1=-3.5N force due to p= 1.77N angle between this forces=180-theta=180-53=127 so mag=sqrt((-3.5)^2+1.77^2-2*3.5*1.77*cos127) =4.77N direction of equivalent force is tan a= 1.77cos37/-3.5+1.77sin37 tan a =-.058 a=-30.13 so 30 deg bellow axis outward

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