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Capacitor Problem!

  • Thread starter sahil_time
  • Start date
  • #1
108
0
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?
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31
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).
 
  • #4
108
0
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 cant be lost as heat...without resistance!
 
  • #5
1,506
17
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)
 

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