The grey charge distribution shown generates an electric field corresponding to
ID: 1540767 • Letter: T
Question
The grey charge distribution shown generates an electric field corresponding to the following equipotential surfaces. the potentials at points A and B are V_A = 30 V and V_s = 1.0 V. On this diagram sketch me of the electric held lines resulting from the charge distribution. is the charge distribution positive or negative? (Yes, you have enough information to tell.) Where the electric field strongest? Explain. How much work would take to move a Q = 0.50C point charge along a straight line from B to A? Now consider a semicircular path from B to A. To move the Q = 0.50 C charge along this path, would it take mow work. less work, or the same work, as compared to part (c)? Explain. Which takes more work: Moving charge Q from point C to point A. or moving it from point B to point A? Justify your answer.Explanation / Answer
(a)
The electric field has direction from the high potential region to low potential region. The electric field lines starts from the positive charge and ends at the negative charge. The potential at point A is higher than the point B. so, the charge distribution is negative.
(b)
The density of the field lines represents the strengh of the field. The field is strongest at higher density of lines. so, the electric field is strongest at region C.
(c)
The work done required is,
W= (VA-VB)*q
= (3.0 -1.0)*0.50 = 1 J.
(d)
the work done only depends on the initial and final potentials. so, the work done will be same.
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