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Capacitor concept

  • Thread starter ChrisLM
  • Start date
9
0
1. Homework Statement
It's not a typical question but is a challenging question from my sir, That's consider 2 cases:
1- A capacitor connected with a d.c power supply
2- A capacitor (originally charged) with two ends open.

These two cases undergo two changes,1- increase in the plate-to-plate separation.(d)2- decrease in the superimpose area.(A)

In the first case, potential difference across the capacitor is constant
It's given that when d is increased, energy stored in capacitor would decrease,the difference in energy go back to the power supply. In the above process, is any energy needed?(by external force to increase d?) The answer is yes but I can't realise what has happen.
Also, when A is decreased, The energy stored aggain decreases and go abck to the power supply.However, this time no energy is required. Why?

Similarly, for the second case, these two changes would result in an increase in the energy stored in the capacitor, and both of them does requires energy to undergoes changes 1 and 2.
2. Homework Equations
C=Q/V



3. The Attempt at a Solution
 

Answers and Replies

454
0
In all cases it takes energy to pull apart the plates, decreasing the area takes energy too. Any method I can think of reducing the area will bring the charges further apart. if the capacitor is connected to the power supply they flow away through the power supply (the energy is probably lost). If the charges can't flow away the potential must go up. If the area decreases, the charges will flow to the parts of the plates that are still next to each other and the field will increase, if the distance increased, the field will stay the same (if d^2 is still small compared to A). In both cases the potential, wich is the field, integrated along a path from one plate to the other will go up.
 

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