A proton is placed in an electric field of intensity 700 N/C. What is the magnitude and direction of the acceleration of this proton due to this field?

Question

Quizplus · Accepted Answer

The acceleration experienced by a charged particle in an electric field is given by the formula:
a = qE/m
where a is the acceleration, q is the charge of the particle, E is the electric field intensity, and m is the mass of the particle.
In this case, we have a proton, which has a charge of +1.6 × 10^-19 C, and a mass of 1.67 × 10^-27 kg.
Substituting these values and the given electric field intensity, we get:
a = (1.6 × 10^-19 C)(700 N/C)/(1.67 × 10^-27 kg) = 6.71 × 10^10 m/s^2
The direction of the acceleration will be in the direction of the electric field because the proton has a positive charge and will be attracted towards the negative end of the field. Therefore, the correct answer is (C).

A Proton Is Placed in an Electric Field of Intensity