Question

If 2.0×10^−4 C of charge passes a point in 2.0×10^−6 s , what is the rate of current flow?1.0×10^−10 A1.0×10^2 A4.0×10^−1 A4.0×10^1 A

Answer 1

This problem is about the rate of the current. It's important to know that refers to the quotient between the electric charge and the time, that's the current rate.

`I=(Q)/(t)`

Where Q = 2.0×10^−4 C and t = 2.0×10^−6 s. Let's use these values to find I.

`\begin{gathered} I=(2.0*10^(-4)C)/(2.0*10^(-6)\sec ) \\ I=1.0*10^(-4-(-6))A \\ I=1.0*10^(-4+6)A \\ I=1.0*10^2A \end{gathered}`

As you can observe above, the division of the powers was solved by just subtracting their exponents.

Therefore, the rate of the current flow is 1.0×10^2 A.