An electron is placed in an xy plane where the electric potential depends on x a

Question

An electron is placed in an xy plane where the electric potential depends on x and y as shown in the figure (the potential does not depend on z). The scale of the vertical axes is set by Vs = 500 V. In unit-vector notation, what is the electric force on the electron?

Chapter 24, Problem 039 An electron is placed in an xy plane where the electric potential depends on x and y as shown in the figure (the potential does not depend on z). The scale of the vertical axes is set by Vs = 500 V. In unit-vector notation, what is the electric force on the electron? 0.4 0 0.4 x (m) y (m) j Units Number

Explanation / Answer

The electric field (along some axis) is the (negative) derivative of the potential V with respect to the corresponding coordinate. In this case, the derivatives can be read off of the graphs as slopes (since the graphs are of straight lines). Therefore,

Ex = – dV d x = – (-500 V)(0.20 m) = 2500 V/m = 2500 N/C

Ey = – dV d y = – (300 V)(0.30 m) = –1000 V/m = –1000 N/C

so, Electric field, E = (Ex2 + Ey2) = 2692.58 ~ 2693 N/C

and direction, theta = tan-1(Ey/Ex) = -21.8°

The force on the electron is given by where q = –e. The minus sign associated with the value of q has the implication that F points in the opposite direction from E.

E (which is to say that its angle is found by adding 180º to that of E ). With e = 1.60 × 10–19 C, we obtain

F = ( -1.60 x 10-19 C)[(2500 N/C)i - (1000 N/C)j] = (-4.0 x 10-16 N)i + (1.60 x 10-16 N)j

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