Cutnell, Physics, 10e Help I System Announcements hapter 18, Problem 78 y touch

Question

Cutnell, Physics, 10e Help I System Announcements hapter 18, Problem 78 y touch The charges on three identical metal spheres are-18.0 C, 6.00 IC, and 3.00 pc. The spheres are brought together so they ach other. They are then separated and placed on the x and y axes, as shown in the figure. Treat the spheres as if they were partides. What is the net force ((a) magnitude and (b) direction) exerted on the sphere at the origin? Express the direction as an angle in the range (-180, 180°1 with respect to the positive x-axis direction. ty 42 3.20mm 43 3.20mm (a) Number (b) Number Units Units Click if you would like to Show Work for this question: Open Show Work Question Attempts: 0 of 5 used SAVE FOR LATE

Explanation / Answer

The net charge on each of the three spheres when they are brought together,

Q = [ (-18) + 6 + 3 ] / 3 = -3 C = -3*10^-6 C

Basically q1 = q2 = q3 = Q = -3*10^-6 C

The distance between these spheres: d = 3.2 mm = 3.2*10^-3 m

Force on q1 because of q2:

F12 = k*q1*q2 / d^2 = (9*10^9)*(-3*10^-6)^2 / (3.2*10^-3)^2 = 7.91*10^3 N

Similary force on q1 because of q3:

F13 = k*q1*q3 / d^2 = (9*10^9)*(-3*10^-6)^2 / (3.2*10^-3)^2 = 7.91*10^3 N

(a) The magnitue of the resultant force on q1 at origin,

F = sqrt(F12^2 + F13^2) = sqrt[(7.91*10^3)^2 + (7.91*10^3)^2] = 11.186*10^3 N

(b) The direction of the resultant force on q1 at origin,

  = (- 90°) + (- 90°/ 2) = - 135° Or 45°

Please rate my answer, good luck...

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