1. 1. Three equal point charges, each with charge 1.15 ? C , are placed at the v
ID: 2134413 • Letter: 1
Question
1. 1. Three equal point charges, each with charge 1.15?C , are placed at the vertices of an equilateral triangle whose sides are of length 0.700m . What is the electric potential energy U of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.)
Use ?0 = 8.85
Three equal point charges, each with charge 1.15?C , are placed at the vertices of an equilateral triangle whose sides are of length 0.700m . What is the electric potential energy U of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.) Use ?0 = 8.85 times 10?12C2N?m2 for the permittivity of free space. Two stationary positive point charges, charge 1 of magnitude 3.10nC and charge 2 of magnitude 1.80nC , are separated by a distance of 41.0cm . An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges. What is the speed vfinal of the electron when it is 10.0cm from charge 1? A particle with charge 4.80 times 10?19C is placed on the x axis in a region where the electric potential due to other charges increases in the +x direction but does not change in the y or z direction. The particle, initially at rest, is acted upon only by the electric force and moves from point a to point b along the x axis, increasing its kinetic energy by 6.40 times 10?19J . In what direction and through what potential difference Vb?Va does the particle move?Explanation / Answer
1.
given r=r12=r23=r31=0.7m and
q=q1=q2=q3=1.15uC
U=K(q1q2/r12 +q2q3/r23 +q1q3/q31)
U=3Kq^2/r^2
U=3*(9*10^9)(1.15*10^-6)^2/0.7^2
U=0.07287 J or 72.87 mJ
2.
r11=r21=20.5cm
r12=10cm
r22=41-10=31cm
E1=K1+U1 =-Keq1/r11 -Keq2/r21
E1=-(9*10^9)(1.6*10^-19)[3.1*10^-9/0.205 +1.8*10^-9/0.205]
E1=-3.44*10^-17 J
E2=K2+U2 =-Keq1/r12 -Keq2/r22 +(1/2)mV^2
E2=-(9*10^9)(1.6*10^-19)[3.1*10^-9/0.1+1.8*10^-9/0/31]+(1/2)mV^2
E2=-5.3*10^-17+(1/2)mV^2
E1=E2
-3.44*10^-17 =-5.3*10^-17+(1/2)mV^2
(1/2)mV^2=1.86*10^-17
(1/2)(9.11*10^-31)*V^2=1.86*10^-17
V=6.39*10^6 m/s
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