Question

Two very large charged parallel metal plates are 8.50 cm apart and produce a uniform electric field of 3.05×10⁶ N/C between them. A proton is fired perpendicular to these plates, starting at the middle of the negative plate and going toward the positive plate.

How much work has the electric field done on this proton by the time it reaches the positive plate?

Answer in Joules.

Answer 1

We can calculate work done by force times distance or by change in energy. Either one could be used in this case.

The change in energy is the potential energy difference between the two plates for this charge. This is easy to calculate if you know the potential difference between the plates, but in this case we are only given the field and the plate separation. Of course N/C is the same as V/m so the voltage difference between the plates is (3.05 x 10^6 V/m * 0.085m) = 2.59×10^5V. The charge of the proton is 1.6×10^-19C,

so the work done is -(2.59×10^5 * 1.6×10^-19C) = -4.148 x 10^-14 J

[Force times distance gives exactly the same result since the force is the field strength times the charge, and then multiplied by the plate separation. Note that the result does not depend on the initial velocity of the proton unless relativistic effects come into play.]