Can somebody tell my where I am going to wrong.

  • Context: Undergrad 
  • Thread starter Thread starter dbernat32
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on deriving the formula W=1/2(V^2)C, which represents the work done in charging a capacitor. The user, dbernat32, attempts to manipulate various equations related to capacitance and energy but struggles with the derivation. A key insight provided is the necessity of using integrals to account for the changing voltage as charge is added to the capacitor, leading to the conclusion that the total work is calculated through the integral of V(q) with respect to dq.

PREREQUISITES
  • Understanding of basic electrical concepts such as capacitance, voltage, and charge.
  • Familiarity with the formula W=1/2(V^2)C for energy in capacitors.
  • Knowledge of calculus, specifically integrals, to derive energy equations.
  • Basic principles of electrical circuits and power calculations.
NEXT STEPS
  • Study the derivation of energy stored in capacitors using integrals.
  • Learn about the relationship between charge, voltage, and capacitance in more detail.
  • Explore the concept of work done in electrical systems through calculus.
  • Review the fundamentals of electrical energy storage and its applications in circuits.
USEFUL FOR

Students of electrical engineering, physics enthusiasts, and anyone looking to deepen their understanding of capacitor energy storage and the mathematical principles behind it.

dbernat32
Messages
1
Reaction score
0
W=work(energy) in joules
C=capacitance in farads
A=Amperes
V=Voltage
Q=charge in coulombs
s=seconds
P=power in watts

I know that W=1/2(V^2)C is the correct formula, but I don't know how to derive it.

I am doing the following: C=Q/V, C=As/V, VC=VAs/V, VC=Ps/V, VC=W/V, W=C(V^2)
What am i missing?

--dbernat32
 
Physics news on Phys.org
dbernat32 said:
W=work(energy) in joules
C=capacitance in farads
A=Amperes
V=Voltage
Q=charge in coulombs
s=seconds
P=power in watts

I know that W=1/2(V^2)C is the correct formula, but I don't know how to derive it.

I am doing the following: C=Q/V, C=As/V, VC=VAs/V, VC=Ps/V, VC=W/V, W=C(V^2)
What am i missing?

--dbernat32

Don't know if you studied integrals.
When you add a small charge dq to a condenser at potential V and charge q, you have to make the work V*dq. But now V is not the same anylonger because V = q/C and q is now different. So you have to write V(q) and dW = V(q)*dq.
To get the total work you have to sum all these infinitesimal quantities, that is you have to compute the integral:
Integral(0;Q) V(q)*dq = Integral(0;Q) (q/c)*dq = (1/2)Q^2/C = (1/2)CV^2.
The equality coloured in blue requires knowledge of integrals.
 
Last edited:

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K