value of heat dissipiated

Question

A parrallel plate capacitor of capacitance 0.1 μF . Its two plates are given charge 2μC and 1μC . Prove that value of heat dissipated(in μJ) after switch is closed is 5

Answer 1

Hi Samagra!

 

   As I mentioned, I am not getting their answer.  The way I attacked this problem was to look at the change in energy between the initial state of this system and the final state.  If the final energy state is lower, then that would correspond to the energy dissipated, which would occur as the excess positive charge on the two plates rearranged itself once the switch was closed.

 

  To calculate the energy (U) between the plates, I used:

 

U = (QV)/2

 

   The reason I chose this expression was that it applied even though the two plates are not exactly storing energy like a capacitor.  Both plates have an excess positive charge, so this is never a situation where you have +Q on one plate and -Q on the other, as in a capacitor.  The above expression for energy still applies, though, as it looks at the situation of bringing each new charge (whatever its sign may be) from infinity onto the plates one at a time.  The first charge takes no work, because no other charge is there.  The last charge is brought in against the final voltage between the plates, V, and so the work needed to bring it on is QV, where Q is the total charge on the plates and V is the voltage between the plates with all the charge on.  Since the voltage increases linearly with charge, the average work needed to put the charge on the plates (and thus the energy stored) is the average of the initial and final work (energy) needed to put the charge on: (0 + QV)/2 = QV/2.

 

   The other expressions for capacitor energy, U = (1/2)CV² and U = Q²/2C, assume that the relationship between energy, charge, and voltage on the capacitor is C = Q/V, but that is not true is there is excess charge imposed on the plates, as in this problem.  Hence, I stick with U = QV/2.

 

   We know the total charge on the plates.  We also know the voltage between the plates once the switch is closed:  5 V (since it is then held by the battery).  The challenge then is determining the magnitude of the voltage between the plates in the initial state, with the switch open.

 

   To determine that, I used the expression:

 

|V| = ∫ E · ds

 

   integrated along a line from one plate to the other.  I do not need to use this to determine the polarity with the switch open.  I know the plate with the larger charge is at the higher potential, since all the charge is positive.

 

    To use this, I need first to determine an expression for the electric field between the plates.  In most capacitors, the distance between the plates is much smaller than the area of the plates, so we can approximate the plates as infinite conducting plates with the excess charge spread uniformly across them.  The charge density, σ, on each plate is thus (Q/A).  The expression for an electric field from an infinite conducting plate is:

 

E = σ/ε_o  (where ε_o is, as always, the (constant) permittivity of free space)

 

I will add more later.  I am reconsidering some of the work I did before.