Please show your work 1.) A moving particle encounters an external electric fiel

Question

Please show your work

1.) A moving particle encounters an external electric field that decreases its kinetic energy from 10270 eV to 6700 eV as the particle moves from position A to position B. The electric potential at A is -56.5 V, and the electric potential at B is +26.1 V. Determine the charge of the particle. Include the algebraic sign (+ or ?) with your answer.
____________C

2.) The drawing below shows a graph of a set of equipotential surfaces in cross section. The grid lines are d = 5.5 cm apart. Determine the magnitude and direction of the electric field at position D. Specify whether the electric field points toward the top or the bottom of the drawing.

Explanation / Answer

a)

here by using

Delta E = q Delta V

10270 - 6700 = q( -56.5 - 26.1)

q = (10270 - 6700) / ( -56.5 - 26.1)

q = -43.22 e

then q = -43.22 * 1.602 * 10^-19 = - 69.23 * 10^-19 C

q = -6.92 * 10^-18 C

so the charge is -6.92 * 10^-18C

b)

Notice the equipotentials above D and below D are flat and parallel. You can infer that all the equipotentials inbetween are also flat and parallel. This means the change in potential is uniform in the neighborhood around D and therefore the Electric field is constant in this region. So in the formula (Vf-Vi)/ds you can use any pair of voltages in the neighborhood of D ,with the appropriate ds, to find E.
So the ones given are 450 V & 550 v and they can be used;

E = (550 - 450) / 0.11 = 909.09 V/m

For the direction; the E-field is always perpendicular to the equipotentials and points toward decreasing potential. So in your case it would point down.

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