# Why energy is lost in capacitor if wires have no resistance?

1. Jan 7, 2016

### Dexter Neutron

The question in textbook:-
A 4uF capacitor is charged by connecting it to 200V supply.It is then connected to an uncharged 2uF capacitor.How much electrostatic energy of the first capacitor is lost in form of heat and electromagnetic radiation?

My question is that why the energy must be lost if the wires in the circuit are assumed to be of 0 resistance.
The energy must simply be divided according to the capacitance of both the capacitor and remain conserved in the system.

2. Jan 7, 2016

### A.T.

Accelerated charges radiate EM waves.

3. Jan 7, 2016

### jbriggs444

It is a bit like the conundrum of an irresistible force and an immovable object. If you try to reconcile them then something has to give. Either the force was not actually irresistible or the object was not actually immovable.

If you connect an ideal charged capacitor through ideal wires to an ideal uncharged capacitor then you have a non-zero potential difference across a zero resistance. That is a forbidden condition. Some component must not actually be as ideal as we had supposed. [Alternately, one could consider that an infinite current across a zero resistance for an infinitesimal time leads to an indeterminate energy loss, but let us not go there].

Real wires have inductance. That inductance will limit the rate at which current can increase in the wire for a given potential difference. It will also cause the circuit to "ring". The current will not stop flowing at the equilibrium point when both capacitors have the same potential difference. Inductance will cause the current to keep flowing past the equilibrium point. Eventually the current will stop and reverse and the cycle will continue.

Ideally, this would keep going forever. However, if the resistance in the wires is not quite zero then the oscillation energy will decay into heat. If the inductance is not ideal then the oscillation energy will also decay into electromagnetic radiation. No real wires, capacitors or inductors are ideal. Regardless of the details of the actual situation, decay is inevitable and the energy will be lost one way or another.

4. Jan 7, 2016

### lychette

there will be a (rapidly) changing current and a changing current produces electromagnetic radiation...this could be detected as a pulse of radio waves.

5. Jan 7, 2016

6. Jan 7, 2016

### Svein

What the textbook expects: Let C1 be 4μF, C2 be 2μF and V1 be 200V. Then the charge on C1 will be C1V1 = 4E-6⋅200c = 8E-4c. After the capacitors are connected and everything is stable, the voltage across them will be Q/(C1+C2) = 8E-4/6E-6V=133V. Now the energy before connecting the capacitors is ½C1V12= 0.08J and the energy after is ½(C1+C2)V22=0,053333J. The difference is lost in heat, radiation etc.

7. Jan 9, 2016

### Dexter Neutron

But this isn't true for electron then how could energy be lost.

8. Jan 9, 2016

### Staff: Mentor

What do you mean? Accelerating electrons is exactly how radio transmitters work.

9. Jan 9, 2016

### Dexter Neutron

I mean that I studied in rutherford's atomic model's drawback that electrons dont release energy while they are constantly being accelerated in their motion.

10. Jan 9, 2016

### Staff: Mentor

We aren't talking about electrons in their atomic orbitals, but about free electrons which are moving throughout the conductor. These electrons do indeed radiate energy when accelerated.

11. Jan 9, 2016

### Dexter Neutron

Thanks.

12. Jan 9, 2016

### Staff: Mentor

This is a purely classical problem and thinking about electrons does not help, in my opinion. I think one big pedagogical problems in learning EM is the tendency to jump to explanations based on electrons when not needed.

Just focus on charge and current. The nature of the charge carriers is rarely important classically.