A charge of 1.5 mu C is distributed uniformly over the surface of a solid conduc

Question

A charge of 1.5 mu C is distributed uniformly over the surface of a solid conducting sphere of radius R-6 cm, Point A is a distance of 3.0 cm from the center of the sphere, and point B is a distance of 90 cm from the center of the sphere. The potential difference |V_B - V_A| is: 75 kV 150 kV 225kV 37.5 kV The speed of an electron that is accelerated from rest through a potential difference of 728.8V is: a) 8.0 times 10^6 m/s b) 16 times 10^6 m/s c) 12 times 10^8 m/s d) 24 times 10^6 M/s Consider three point charges at the vertices of an equilateral triangle as shown in the figure. The total electrical potential energy of the system, (with the zero reference at infinity) is: a) -0.99 J b) -0.81 J c) -1.26 J d) -1.485 J

Explanation / Answer

VA = (kQ/2R)(3-(r/R)2)

=(9*10^9 *1.5*10^-6/2*0.06)(3-(0.03/0.06)2)

=371250V

VB = kQ/r=9*10^9*1.5*10^-6/0.09=150000 V

difference=371250-150000=225 kV

2)0.5mv2=eV=1.6*10^-19*728.8

v=16*10^6 m/s

2)k[q1q2+q2q3+q3q1]/r=-9*10^9*[6*12-6*3-3*12]*10^-12/0.2=-0.81J

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