The lines show the equipotential contours in the plane of three point charges, Q
ID: 2168006 • 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 work required to move a charge of -0.31?10-12C from i to b.
answer: 1.86*10^-9J
Calculate the size of the electric field at g .
answer:?
Calculate the size of the force on a charge of 9.60?10-19 C at k.
answer:?
Calculate the size of Q3. The magnitudes of the three charges are in the exact ratios of 1 to 2 to 3.
answer:?
Explanation / Answer
The potential due to a point charge is V(r) = K*q/r. For a given distance r from a charge, the lower potential means a lower charge. This means a given potential (say 5 kV) is farther away from a lower charge. Look at the 5 kV lines near the charges; It is farthest from Q3, so the relative value of Q3 = 1. Next closest is Q1, which then has a relative value of 2, and finally, Q2 has the relative value of 3. Set the unkown value of Q3 = Q. Then the potential a distance r from Q3 is K*Q/r. The potential at a distance r from Q1 is 2*K*Q/r, and from Q2 it is 3*K*Q/r. Pick a point on the 2 kV potential line that is roughly equal distance from each charge. Measure the distance from that point to each charge and compute that charge's contribution to the potential at that point from the potential formulas for each charge given above, add them up (take the signs into account: Q3 will be +, the others -) and set that = to 2 kV and solve for Q, which is the charge on Q3. It appears the 1 kV line is one that is easier to find an equidistant point on. You want roughly equal distances so that any error in distance measurement has roughly equal affect on each charge's potential. Picking a point too close to one charge makes that charge's potential computation more sensitive to measurement errors.
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