This is a 3 part question, help would be very much appreciated. For the last que

Question

This is a 3 part question, help would be very much appreciated. For the last question, the answer must be within +- 5.0%

Two point tharges fixed kealions pro duer an elertrie field nas shown. low would a negalive charge placed al point X mown? 1. Along an equipotential plane 2. Toward charge A. 3. Tuward charge B l006 (part of 2) 10.0 points The electric field at point Xis 1. wenker Lhan the nu point Y. 2, stronger than the field at point Y. 3. the same res that the field point Y 007 (part 3 of 3) 10.0 points Eestinale the ralio ol the magniuude ol chargn A to the mRgnitude of eharge R.

Explanation / Answer

part(1)

we know that the charge of each point-charge is positive or negative. In electric field line drawings, positive charges are sources for electric field lines (lines point away) and negative charges are sinks (lines point towards them). Opposite charges attract. so placing a negative charge on point X would make it move toward charge B, because B is a positive test charge.

part(2)

we know that electric field is

E = k*q/r^2 it depend on distance betweeb two charges.

the electric field at point X would be stronger than the electric field at point Y, because it is closer to the center, where the force is the strongest (and shortest distance) between the two.

part(3)

A source with more field lines coming off it indicates that it has a stronger charge.

A= 13 lines ; B = 9 lines

so

ratio of A/B = 13 / 9 which is the ratio of the magnitude of the charges.

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