Capacitor Problem: Work Done by a Battery in Charging

[sahil_time]

I came across a question in my Physics Paper
"What is the work done by a battery in charging a capacitor" (Assume circuit has no resistance).

This is what i thought...
The Energy in the field between the capacitor plates is 0.5CV^2 where C=capacitance and
V=Battery EMF.So the ans should be 0.5CV^2.

BUT the actual solution in paper was...
Net Work done=Net Charge * Voltage
W=QV
W=CV^2 .

What is the correct ans?

 

[rock.freak667]

The energy supplied by the battery is E = QV and Q=CV to get E=CV²

the energy stored by the capacitor is E=½ CV² (if I remember correctly, the other ½ is lost as heat).

 

[256bits]

Maybe this will help

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html

 

[sahil_time]

The energy supplied by the battery is E = QV and Q=CV to get E=CV²

the energy stored by the capacitor is E=½ CV² (if I remember correctly, the other ½ is lost as heat).

It can't be lost as heat...without resistance!

 

[technician]

The energy supplied by the battery is Q x V. The energy stored on the Capacitor is 0.5Q x V.
Energy can be lost by
1) Resistance in the connecting wires
2) Sparking when the switch is closed
3) Electro-magnetic radiation from the connecting wires as the charging current flows. The flowing charge is a changing current and a changing current produces electro-magnetic (radio) radiation
In the absence of 1) and 2) there is always 3)