Question

A uniform electric field of 8 V/m exists between the plates of a parallel plate capacitor. How much work is required to move a 20 mC point charge from the negative plate to the positive plate if the plate separation is 0.050 m

Answer 1

Hello!

Recall the equation for the force experienced by a charged particle in an electric field:

`F_E = qE`

q = Charge of particle (20 mC)
E = Electric field strength (8 V/m)

We also know that 'Work' is equivalent to the following:

`W = F \cdot d`

W = Work (J)

F = Force (N)

d = distance (between the plates in this instance, 0.050 m)

If we substitute 'qE' for 'F':

`W = qEd`

Plug in the given values and solve.

`W = (0.02)(8)(0.05) = \boxed{0.08 J}`

**Since we are pushing a positive particle towards the positive plate, we are increasing its electric potential energy, so WE are doing positive work on the particle.