The lines show the equipotential contours in the plane of three point charges, Q
ID: 2221278 • Letter: T
Question
The lines show the equipotential contours in the plane of three point charges, Q1, Q2, and Q3. The values of the potentials are in kV as indicated for the +5, 0, and -5 kV contours. The positions of the charges are indicated by the dots.
Calculate the size of Q3. The magnitudes of the three charges are in the exact ratios of 1 to 2 to 3.
The ratio 1:2:3 does not imply that Q1=1:Q2=2:Q3=3 You have to find out which charge is the smallest/largest from the plot Pick a point on the +5kV contour (or even better on the +2kV contour). Measure the distance to the 3 charges (in m). Calculate the potential assuming charges of Q, 2Q and 3Q (watch the signs) Note that 2Q refers to Q1 and Q refers to Q2, etc. Set the potential equal to 5 kV and solve.
Explanation / Answer
As seen from plot, Q3 > Q1 > Q2 taking Q3 = 3q, Q1 = 2q and Q2 = q Q1 at 2 , Q2 at 6 and Q3 at 5 picking a point aboveQ3 on 5kV contour distance fro Q3 = r =(taking Q3 at height of 1.5 cm and point at height of 2.5 cm) = 2.5 - 1.5 = 1 cm = 0.01 m since electric potential,V = kQ/r where k = 9*10^9 5000 V = 9*10^9*3q/0.01 => q = 1.862 nC thus Q3 = 3q = 5.56 nC Ans
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