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

AI Thread Summary
A DC current of 3 A flows through an initially discharged capacitor, and the question is about the electrical charge after 2 microseconds. The equation for charge is given as q(t) = integral of i(t) dt from 0 to t. The initial attempt at solving this incorrectly used (3t^2)/2, leading to confusion about the integration of a constant current. The correct approach should recognize that integrating a constant current results in a linear function of time, not a quadratic one. Clarification on the integration process is essential for accurate calculations.
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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