The lines show equipotential contours in the plane of three point charges, Q 1 ,
ID: 1542163 • Letter: T
Question
The lines show equipotential contours in the plane of three point charges, Q1, Q2, and Q3. The positions of the charges are marked by dots. The values of the potentials are in kilovolts as indicated, e.g., +5 kV, 5 kV; the contour interval is 1 kV. The letters denote locations on the contours. For distances use the scale below the picture, to an appropriate accuracy. Select all the correct statements, e.g., AB.
A) Charge Q3 is the largest positive charge.
B) Charge Q2 is the largest negative charge.
C) Q3 is a negative charge.
D) The electric field at a is stronger than at d.
E) The electric field at g is zero.
F)The force on a proton at g points to the top of the page.
2) Calculate the work required to move a charge of -3.20×10-12 C from d to k.
3) Calculate the magnitude of the electric field at f.
Q2 g o/volts rn cm 2 4 Q 1Explanation / Answer
A. the potentials at points i and b are +3kV and -3kV respectively. to move the charge from i to b,
work done = charge x potential difference = -0.67×10^-12C x -6x10^3 V = 4.02 x 10^-9J = 4.02 nJ.
B. the perpendicular distance between the equipotentials either side of g (-1kV and +1kV) is 9mm.
(to measure distance XY, place the straight edge of a sheet of paper against X and Y, mark the positions of X and Y on the edge of the paper, and measure this distance using the scale given in the figure.)
electric field strength at g = PD/distance = 2x10^3V/9x10^-3mm = 0.22x10^6 V/m.
C. the perpendicular distance between the equipotentials either side of k (-1kV and +1kV) is 12.5mm.
electric field strength at k = PD/distance = 2x10^3V/12.5x10^-3mm = 0.16x10^6 V/m.
force on charge at k = charge x electric field strength = -3.20×10-12 C x 0.16x10^6 V/m = 7.7x10^-14 N.
D. (THE DIFFICULT PART.) to find the charge Q3 we can use the SUPERPOSITION PRINCIPLE : the electrostatic potential at any point in the diagram is the sum of the potentials from all 3 charges. the potential at distance r from a charge Q is Q/4r, where is the permittivity of free space. we pick any convenient point X on an equipotential and measure the distances r1, r2,r3 of Q1, Q2, Q3 from X. then
(1/ 4)(Q1/r1 + Q2/r2 + Q3/r3) = P
where P = potential at X.
(comment: if required, we could use this equation to find all 3 charges Q1, Q2, Q3 by selecting 3 different points X, Y, Z then solving 3 simultaneous equations. alternatively, we could measure distances from several points and average the resulting values of Q3 to improve accuracy.)
judging by the distance from each charge to the nearest equipotential (+5 or -5 kV), Q3 has the largest magnitude (because Q/r is the same for each charge, so large distance r to the equipotential means Q must also be large) and is +ve. Q1 is comes in between and is -ve, and Q2 is smallest, also -ve.
so Q3 : Q1 :Q2 = +3 : -2 : -1. if Q2 = -q then Q1 = -2q and Q3 = +3q. substituting above,
q(-2/r1 -1/r2 + 3/r3) = 4P.
for accuracy, choose X as far from each charge as possible. X cannot be on the P= 0 V equipotential, because the equation would then give us q=0 for all values of r. i choose X as the point where the +1kV equipotential meets the left lower edge.
distances measured to X from Q1, Q2, Q3 are 62, 70, 35.5 mm respectively.
P= +1kV, = 8.85x10^-12 F/m, 4P = 1.1x10^-7 C/m.
(-2/r1 -1/r2 + 3/r3) = (-2/0.062 -1/0.070 + 3/0.0355) = 37.96 m^-1.
q = 1.1x10^-7 / 37.96 = 2.92 x 10^-9 C.
ANSWER: Q3 = 3x (+2.92nC) = +8.8 nC.
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