4. In a vacuum, two particles have charges of q 1 and q 2 , where q 1 = +3.3 In

Question

4. In a vacuum, two particles have charges of q1 and q2, where q1 = +3.3

In a vacuum, two particles have charges of q1 and q2, where q1 = +3.3 mu C. They are separated by a distance of 0.29 m, and particle 1 experiences an attractive force of 3.0 N. What is q2 (magnitude and sign)? Two point charges are fixed on the y axis: a negative point charge q1 = -27 mu C at y1 = +0.24 m and a positive point charge q2 at y2 = +0.33 m. A third point charge q = +8.7 mu C is fixed at the origin. The net electrostatic force exerted on the charge q by the other two charges has a magnitude of 28 N and points in the +y direction. Determine the magnitude of q2. n Multiple-Concept Example 9 you can use the concepts that are important in this problem. A particle of charge +11 mu C and mass 3.90 times 10-5 kg is released from rest in a region where there is a constant electric field of +510 N/C. What is the displacement of the particle after a time of 1.68 times 10-2 s? Two point charges of the same magnitude but opposite signs are fixed to either end of the base of an isosceles triangle, as the drawing shows. The electric field at the midpoint M between the charges has a magnitude EM. The field directly above the midpoint at point P has a magnitude EP. The ratio of these two field magnitudes is EM/EP = 5.31. Find the angle? in the drawing. A small plastic ball with a mass of 6.55 times 10-3 kg and with a charge of +0.158 mu C is suspended from an insulating thread and hangs between the plates of a capacitor (see the drawing). The ball is in equilibrium, with the thread making an angle of 30.0 degree with respect to the vertical. The area of each plate is 0.0165 m2. What is the magnitude of the charge on each plate?

Explanation / Answer

4)let the charge be q uC.so,

F=k*q1*q2/r^2

or 3=9*10^9*3.3*10^-12*q/0.29^2

or q= -8.49 uC

12)let the charge density on the capacitor be x.so,

electric field between the plates of the capacitor=x/eo.

so,

tan(30)=qE/mg

or 0.577=0.158*10^-6*x/(8.85*10^-12*6.55*10^-3*9.81)

or x=2.07*10^-6 C/m^2

so charge on each plate-x*area

=2.07*10^-6*0.0165

=0.0342 *10^-6 C

=3.42*10^-8 C

10) force on the particle=q*E

=510*11*10^-6

=5.61*10^-3 N

so acceleration=F/m

=5.61*10^-3/(3.9*10^-5)

=143.846 m/s^2

so distance covered=0.5at^2

=0.5*143.846*(1.68*10^-2)^2

=0.02 m

6) let the charge q2 be x.so,

F=kq1q2d^2

or 28=9*10^9*8.7*10^-12*(27/0.24^2 - x/0.33^2)

or x=12.104 uC

so the charge q2 is 12.104 uC

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