Why Does Removing the Negative Sign in the Charge Equation Affect the Result?

[cseet]

Hi all,

I'm stuck on this charge equation:

Q=Q * e (-t/RC)
= 780 * e (-9.8/3.3)
I've got 40.03
but the answer is 15200

if I took away the negative sign from the equation then I'll be able to get 15200. but that should not be the way isn't it? can somebody advise me how should I go about with this?

thanks
cseet

 

[arildno]

I'd say your answer is correct; then the amount of net charge is depleted exponentially from an initial level Q_{0}=780.
It really depends on what your original equation is trying to model.

 

[M98Ranger]

Hi cseet,

Thank you for reaching out for help with the charge equation. It looks like you are struggling with the negative sign in the exponent. Let me try to explain it in a simpler way.

The charge equation is used to calculate the amount of charge (Q) in a capacitor after a certain amount of time (t) has passed. The variables in the equation are Q, t, R, and C. The letter "e" represents the mathematical constant, Euler's number, which is approximately equal to 2.71828.

Now, looking at your calculations, you are correct in saying that Q=780 and t=9.8. However, the negative sign in the exponent is causing confusion. In mathematics, a negative exponent means that the number should be divided by itself a certain number of times. In this case, it means that e should be divided by itself 9.8/3.3 = 3 times.

So, the calculation should look like this:

Q=780 * e^(-9.8/3.3)
= 780 * (e/e/e) (since e/e/e = 1)
= 780 * (1/1/1) (since e = 1)
= 780 * 1 (since 1/1/1 = 1)
= 780

Therefore, the correct answer is 780, not 40.03 or 15200. I hope this helps clarify your confusion. Remember to always pay attention to the negative sign in exponents and follow the correct order of operations in your calculations. Keep practicing and you will get the hang of it. Good luck!