RC Time Constant Question (Easy)

[beanus]

Homework Statement

A 12.8 micro-F capacitor is connected through a 0.890 M-ohm resistor to a constant potential difference of 60.0 v.

Homework Equations

q=CE(1-e^(-t/(RC))
i=(E/R)-(q/(RC))

The Attempt at a Solution

Compute the charge on the capacitor at the following times after the connections are made: 0 s, 5.0 s, 10.0 s, 20.0 s, and 100.0 s.

Solved for q and according to masteringphysics (the homework program we use) I got all the q values correct. Here they are:
0 , 2.7e−4 , 4.5e−4 , 6.4e−4 , 7.7e−4


Second part, and this is the part that I need help on. I think I'm right but the program says I'm incorrect on the fourth term

Compute the charging currents at the same instants. Calculated i:
6.74e-5 , 4.37e-5 , 2.79e-5 , 1.12e-5 , -1.76e-7

It says "Term 4: Very close. Check the rounding and number of significant figures in your final answer."

What am I doing wrong?

Thanks!

 

[SammyS]

Probably too much too much round off.

 

[beanus]

Probably too much too much round off.

So am I right?

 

[SammyS]

So am I right?

Close to right.

A small round off in E/R makes a big difference and/or q/(RC) makes a big % difference in E/R - q/(RC)when E/R and q/(RC) are nearly the same size.

The correct answer for current @ 20.0 s is ≈ 11.65 μA .

Your answer for 100 seconds is nonsense, since it's negative.

BTW:

Another way to calculate the current at time, t, is to use $I=(I_0)e^{-t/(RC)}$, where I₀ = E/R

 

[asteriatic]

I would first check my calculations and make sure I am using the correct equations and values. It is possible that there was a rounding error or incorrect use of significant figures in the calculation for the fourth term. I would also double check the units to make sure they are consistent throughout the calculation.

If everything checks out, I would consider the physical meaning of the answer. It is unusual for the charging current to decrease over time, so I would question whether the negative value makes sense in this context. It is possible that there is a mistake in the problem statement or in the given values.

If the issue still cannot be resolved, I would consult with a colleague or the instructor for further clarification or assistance. It is important to always double check and verify results in scientific calculations to ensure accuracy.