PROBLEM Four charged particles are fixed in place at the corners of a square wit

Question

PROBLEM

Four charged particles are fixed in place at the corners of a square with side length 0.5 meters. (The square is oriented with its edges aligned north-south and east-west.) The particles each have a charge of 5.0 nC. An electron, free to move, is introduced 0.4 meters to the east from the center of the square. Address the following questions:

a) What is the electric field at the location of the electron? Give magnitude and direction.

b) What is the electrostatic potential at the location of the electron?

c) What is the electrostatic force on the electron? Give magnitude and direction. (Hint: apply one or more of yourr esults from above, rather than starting this calculation from scratch.)

Explanation / Answer

a)The vertical components of fields cancel each other out due to symmetry.

E=Ex=2*kqCos(theta1)/r1^2+2kqCos(theta2)/r2^2=2*9*10^9*5*10^-9*(0.15/(sqrt(0.15^2+0.25^2)*(0.15^2+0.25^2))+0.65/(sqrt(0.65^2+0.25^2)*(0.65^2+0.25^2)))=718N/C eastward

b)V=2kq/r1+2kq/r2=2*9*10^9*5*10^-9*(1/sqrt(0.15^2+0.25^2)+1/sqrt(0.65^2+0.25^2))=437.9V

c)F=qE=eE=1.6*10^-19*718N/C eastward=1.15*10^-16 N eastward

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