What is the Electrical Charge on a Capacitor After 2 Microseconds?

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Discussion Overview

The discussion revolves around calculating the electrical charge on a capacitor after a specific time interval when a constant direct current flows through it. The focus is on the mathematical formulation and integration involved in determining the charge based on the current supplied.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant states the equation for charge as q(t) = integral i(t) dt between t=0 and t, suggesting that the charge after 2 microseconds is (3t^2)/2.
  • Another participant questions the setup of the integral, asking what function was integrated and implying that the integral may have been set up incorrectly.
  • A participant attempts to clarify that the current is constant at 3 A, indicating that the integral of a constant should not yield a t^2 term.
  • One participant acknowledges a mistake in their reasoning and expresses gratitude for the assistance provided.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct setup of the integral for calculating the charge, and there is ongoing clarification regarding the integration of a constant function.

Contextual Notes

There is uncertainty regarding the correct formulation of the integral and the assumptions made about the current function. The discussion does not resolve the mathematical steps necessary for the solution.

patep023
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Homework Statement



A dc current of 3 A flows through an initially discharged capacitor.
After 2 micro s what is the electrical charge on the capacitor?

Homework Equations


q(t)= integral i(t) dt between t=0 and t

The Attempt at a Solution


(3t^2)/2=6x10^-12 C. Is this right?
 
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patep023 said:

Homework Statement



A dc current of 3 A flows through an initially discharged capacitor.
After 2 micro s what is the electrical charge on the capacitor?

Homework Equations


q(t)= integral i(t) dt between t=0 and t


The Attempt at a Solution


(3t^2)/2=6x10^-12 C. Is this right?

Your relevant equation is okay, but it looks like you set up the integral incorrectly. What function did you integrate?
 
f(q)=i ?so integral of 3A which is (3t^2)/2 ?
 
patep023 said:
f(q)=i ?so integral of 3A which is (3t^2)/2 ?

Nope. Your function is a constant. Dropping units for now, I(t) = 3. There's no "t" in the function to become a ##t^2## upon integrating.
 
ye I was just being stupid, ty for the help
 

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