Question

You charge a parallel-plate capacitor, remove it from the battery, and prevent the wires connected to the plates from touching each other. When you increase the plate separation, what happens to the following quantities? (a) C increases decreases stays the same (b) Q increases decreases stays the same (c) E between the plates increases decreases stays the same (d) ΔV

Answer 1

Answer: a) C decreases; b) Q stays the same; c) E is the same

d) ΔV increase

Explanation: In order to explain this problem we have to consider the following:

C=εoA/d where A and d are the area and the separation of the plates, respectively.

Increasing d, produces a decrease of C.

Q remain constant becasuse the plates are charges and the wire are isoloted each other.

We also know that  ΔV=E*d where E is electric field between the plates.

And E= Q/εo*A  ( a constant between the plates)

As we can see  from above, ΔV depends directely of the d so if d increase ΔV also increase. To do that we have to do work on the system.