1.An object has a charge of ? 2.1 ? C. How many electrons must be removed so tha

Question

1.An object has a charge of ?2.1 ?C. How many electrons must be removed so that the charge becomes +2.9 ?C?

2.Two tiny conducting spheres are identical and carry charges of -21.6

An object has a charge of ?2.1 ?C. How many electrons must be removed so that the charge becomes + 2.9 ?C? Two tiny conducting spheres are identical and carry charges of - 21.6 mu C and + 51.5 mu C. They are separated by a distance of 2.33 cm. What is the magnitude of the force that each sphere experiences, and is the force attractive or repulsive? The spheres are brought into contact and then separated to a distance of 2.33 cm. Determine the magnitude of the force that each sphere now experiences, and state whether the force is attractive or repulsive. Two point charges are fixed on the y axis: a negative point charge q1 = - 24 mu C at y1 = + 0.19 m and a positive point charge q2 at y2 = + 0.33 m. A third point charge q = + 8.2 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 24 N and points in the + y direction. Determine the magnitude of q2. Suppose a single electron orbits about a nucleus containing two protons ( + 2e), as would be the case for a helium atom from which one of the two naturally occurring electrons is removed. The radius of the orbit is 2.65 10 - 11 m. Determine the magnitude of the electron's centripetal acceleration. A small spherical insulator of mass 4.55 10 - 2 kg and charge + 0.600 ?C is hung by a thread of negligible mass. A charge of ?0.900 ?C is held 0.150 m away from the sphere and directly to the right of it, so the thread makes an angle ? with the vertical (see the drawing). Find the following.

Explanation / Answer

1. -2.1E-6 + N*1.6E-19 = 2.9E-6

N=3.125E13

2. attractice
F = k q Q/r^2 = 9.0E9*21.6E-6*51.5E-6/2.33E-2^2=18441 N

b) charge will equalize
q = (-21.6+51.5)/2= 14.95 uC

so repulsive

F = 9.0E9*14.95E-6^2/2.33E-2^2=3705 N

3)
so net F = F1 + F2 = 9.0E9*24.0E-6*9.2E-6/0.19^2+9.0E9*q*9.2E-6/0.33^2=24

q=4.08E-5 C

4) F = k q Q/r^2 = m a

a = 9.0E9*2*1.6E-19^2/2.65E-11^2/9.11E-31=7.2E23 m/s^2

5)

sum forces in the y
T cos theta -mg = 0
T = mg/cos theta

sum forces in the x
-T sin theta + k q Q/r^2 = 0
- m g tan theta + k q Q/r^2 = 0
theta = arctan( 9.0E9*0.6E-6*0.9E-6/0.15^2/(4.55E-2*9.81))= 25.82 degrees

b) T = 4.55E-2*9.81/cos(25.82 degrees)=0.496 N

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