Question

The lines show the equipotential contours in the plane of threepoint charges, Q₁, Q₂, andQ₃. The values of the potentials are inkV as indicated for the +5, 0, and -5 kVcontours. The positions of the charges are indicated by thedots.

Q1 aVolts f5 Qa cm

a-Calculate the work performed by an external agent to move a charge of -0.59x10^-12 C from `i' to `b' ?

b-Calculate the magnitude of the electric field at `g'. ?

c- Calculate the magnitude of the force on a charge of 9.6*10^-19 C at `k'.?

d-Calculate the size of Q3. The magnitudes of the three charges are in the exact ratios of 1 to 2 to 3.

Answer 1

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 = -0.59×10^-12C x -6x10^3 V = 3.54 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 = 9.6×10^-19 C x 0.16x10^6 V/m = 1.536x10^-13 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/4πεr, where ε 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/ 4πε)(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) = 4πεP.

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, ε = 8.85x10^-12 F/m, 4πεP = 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