2.In the previous problem, let Q 1 = 2.20 microCoulombs, Q 2 =-3.20 microC, and
ID: 2137388 • Letter: 2
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2.In the previous problem, let Q1= 2.20 microCoulombs, Q2=-3.20 microC, and Q3= 3.50 microC. (Note that Q1 and Q2 are different now.) The distances are a= 1.20 cm and d= 2.80 cm. Calculate the total electrostatic potential energy of the charge configuration.
In the picture below, the 3 charges Q1, Q2 and Q3 are located at positions (-a,0), (a,0) and (0,-d) respectively. (The origin is the point halfway between Q1 and Q2.) Consider the special case where Q1, Q3 greater than zero and Q2 = -Q1. Assume that the zero of potential is at infinity, as is normal for point charges. Which of the following statements are true? (Choose all correct answers e.g. ABD, CDFG). The electric potential at the origin equals Q3/(4*pi*epsilon0*d). The external work done to bring these charges to this configuration (from infinity) was positive. The electric field at the origin points in the positive y direction, away from Q3. The work required to move Q3 from its present position to the origin is zero. The force on Q3 due to the other two charges is zero. If Q3 is released from rest, it will initially accelerate to the right. The electric potential at any point along the y-axis is positive. In the previous problem, let Q1= 2.20 microCoulombs, Q2=-3.20 micro C, and Q3= 3.50 micro C. (Note that Q1 and Q2 are different now.) The distances are a= 1.20 cm and d= 2.80 cm. Calculate the total electrostatic potential energy of the charge configuration.Explanation / Answer
the statements which are true are
A
D
F
G
For the total eletrostatic potential energy of the system, we will need to consider interaction energies between every pair of point charges and then add those energies
i.e., electrostatic potential energy = energy due to electrostatic interaction between Q1,Q2 and Q1,Q3 and Q2,Q3
=k * Q1Q2/2a + k * Q1Q3/ (a^2 + d^2)^1/2 + k * Q2Q3/(A^2 + d^2)^1/2
= -k * 7.04*10^-12/(2*1.20*10^-2) + k * 7.7*10^-12/(3.05 ^ 10-2) -k*11.2*10^-12/(3.05*10^-2)
=k( -2.93*10^-10 + 2.52*10^-10 - 3.67*10^-10 )
=-k * 4.08 * 10^-10
=-9*10^9*4.08*10^-10
= -3.672 Joules (k=1/(4 *pi *episillon0)
{all units have been converted into SI}
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