Question

An electron is released from rest in a unifor electric field and accelerates to the north at a rate of 145 m/s^2. What is the magnitude and direction of the electric field?

Answer 1

Answer:

E = 8.26*10⁻¹⁰ N/C, due south.

Explanation:

• Assuming no other forces acting on the electron than the electrostatic force due to the electric field, we can apply Newton's 2nd law as follows:

       $F = -eE =ma (1)$

• Solving for E, we can find its magnitude as follows:

       $E =\frac{m*a}{e} = \frac{9.1e-31 kg*145m/s2}{1.6e-19C} = 8.26e-10 N/C (1)$

• The direction of the electric field is by definition the one that would take a positive test charge, so if the electron is accelerated to the north, the electric field would exactly oppose to this direction, so it is directed due south.