Why Does Integrating Current Over Time Equal the Initial Charge of a Capacitor?

  • Thread starter Thread starter Puchinita5
  • Start date Start date
  • Tags Tags
    Capacitor
Click For Summary
Integrating current over time equals the initial charge of a capacitor due to the relationship between charge and current, defined as I = dq/dt. The integral of current (Idt) from zero to infinity represents the total charge transferred, which should equal the initial charge (q(0)). The user attempts to derive this using the equation q(t) = Qe^(-t/RC) and notes that as time approaches infinity, the charge approaches zero. However, confusion arises regarding the limits of integration, with the user questioning whether the integral should be from zero to q(0) instead of zero to infinity. Understanding the correct limits and the implications of the negative sign is crucial for resolving the problem.
Puchinita5
Messages
178
Reaction score
0

Homework Statement



Hi. I had to do a lab on charging and discharging capacitors. In my lab packet, my professor asks:
"We know by definition I=dq/dt. So if we calculate the integral Idt (from zero to infinity) = integral dq (from zero to infinity) it should be equal to q(0) (the initial charge of the capacitor). Explain briefly (show some steps in your derivation) why this is the case.

i'm not 100% sure if I'm understanding the question right. But i'll show my attempt

Homework Equations





The Attempt at a Solution



q(t)=Qe^(-t/RC) where Q is the max charge.

so the integral of dq from zero to infinity should be q(infinity)-q(0)...

and since q(infinity) approaches zero, i'll assume it just equals zero...

my problem is, this leaves me with -q(0)...i don't know if the negative sign makes a huge difference in what he's asking, but i feel like it does? Anyone have a clue?
 
Physics news on Phys.org
Limit of dq cannot be from zero to infinity but the the limit of time can be zero to infinity. It should be zero to qo.
 

Thread 'Magnitude of buoyant force in fluids of different densities'

The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?
View full post »

Similar threads

Final Charge of a Capacitor in a Circuit with Capacitors and Resistors
  • · Replies 9 ·
Replies
9
Views
2K
Discharging a capacitor -- Calculate the current as a function of time
  • · Replies 3 ·
Replies
3
Views
2K
Charge on a grounded conducting sphere in a uniform electric field, after ungrounding and movement
  • · Replies 1 ·
Replies
1
Views
1K
Maximum charge on the plates of a capacitor
  • · Replies 2 ·
Replies
2
Views
2K
Capacitor Questions Charging and Discharging
  • · Replies 4 ·
Replies
4
Views
3K
A charge q approaching two stationary charges q1 and q2
  • · Replies 7 ·
Replies
7
Views
1K
A Capacitor with initial charge
  • · Replies 3 ·
Replies
3
Views
3K
How to Apply Derivatives in Physics Problems?
  • · Replies 9 ·
Replies
9
Views
1K
Where Does the Energy Go in an Ideal Capacitor Circuit?
  • · Replies 16 ·
Replies
16
Views
2K
Charge on inner/outer surfaces of two large parallel conducting plates
  • · Replies 11 ·
Replies
11
Views
3K