Please answer it and correct. FOR NUMBER ONE IT\'S NOT 22.57 AND #2 NOT 19.36 TH

Question

Please answer it and correct. FOR NUMBER ONE IT'S NOT 22.57 AND #2 NOT 19.36 THE ONLY RIGHT ANSWER WAS #3

A point charge q -3.4 UC is located at the origin of a co-ordinate Another point charge q2 6.8 uC is located along the x axis system at a distance x 9.6 cm from q 1) What is 2.x the value of the x-component of the force that q1 exerts on q2? 22.57 bmit You currently have 1 submissions for this question. Only 10 submission are allowed. You can make 9 more submissions for this question. charge q2 is now displaced a distance y2 3 cm in the positive y direction. What is the new value for the x-component of the force that q1 exerts on q2? N Submit 19.63 You currently have 1 submissions for this question. Only 10 submission are allowed. You can make 9 more submissions for this question. A third point charge q3 is now positioned halfway between q1 and q2 he net force on Q2 now has a magnitude of F 2,net 6.856 N and points away from q1 and q3. What is the value sign and magnitude) of the charge q3? Submit 1.13 UC You currently have 1 submissions for this question. Only 10 submission are allowed. You can make 9 more submissions for this question.

Explanation / Answer

1) F12,x = k*q1*q2/x2^2

= 9*10^9*6.8*10^-6*3.4*10^-6/0.096^2

= 22.58 N (towards -x axis)

so,

F12,x = -22.58 N <<<<<<<<<--------------Answer

2) F12 = k*q1*q2/(x2^2 + y2^2)

= 9*10^9*6.8*10^-6*3.4*10^-6/(0.096^2 + 0.03^2)

= 20.57 N (attractive force)

angle made by F12 with x axi, theta = tan^-1(y2/x2)

= tan^-1(3/9.6)

= 17.35 degrees

F12,x = -F12*cos(17.35)

= -20.57*cos(17.35)

= -19.63 N <<<<<<<<<--------------Answer

3) 1.13

4) increase its magnitude and keep its sign the same

5) decrease its magnitude and keep its sign the same

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