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

Increasing the separation in a charged parallel-plate capacitor decreases capacitance while keeping charge and electric field the same, and increases voltage.

Step-by-step explanation:

When you increase the plate separation of a charged parallel-plate capacitor that is disconnected from the battery, the following occurs:

• (a) The capacitance (C) decreases because capacitance is inversely proportional to the distance (d) between the plates.
• (b) The charge (Q) on the capacitor stays the same since no additional charge can flow onto the plates after it is disconnected from the battery.
• (c) The electric field (E) between the plates stays the same as it is directly proportional to the charge (Q) and inversely proportional to the area (A), both of which remain constant.
• (d) The voltage (ΔV) across the plates increases because voltage is directly proportional to the product of the electric field (E) and the distance (d).

Answer 2

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.