这是一份bath巴斯大学PH10006/PH10051作业代写的成功案

Question: My old camcorder runs off of a big lead-acid battery that is labeled 12 volts, $4 \mathrm{AH}$. The “AH” stands for ampere-hours. What is the maximum amount of energy the battery can store?
Solution: An ampere-hour is a unit of current multiplied by a unit of time. Current is charge per unit time, so an ampere-hour is in fact a funny unit of charge.
$$
(1 \mathrm{~A})(1 \mathrm{hour})=(1 \mathrm{C} / \mathrm{s})(3600 \mathrm{~s})
$$
Naw $3600 \mathrm{C}$ is a huge number of
Now $3600 \mathrm{C}$ is a huge number of charged particles, but the total loss of potential energy will just be their total charge multiplied by the voltage difference across which they move:
$$
\begin{aligned}
\Delta P E_{\text {etec }} &=q \Delta V \
&=(3600 \mathrm{C})(12 \mathrm{~V}) \
&=43 \mathrm{~kJ}
\end{aligned}
$$

PH10006/PH10051 COURSE NOTES ：

Question: A charged balloon falls to the ground, and its charge begins leaking off to the Earth. Suppose that the charge on the balloon is given by $q=a e^{-\Delta t}$. Find the current as a function of time, and interpret the answer.
Solution: Taking the derivative, we have
$$
\begin{aligned}
I &=\frac{\mathrm{d}}{\mathrm{d} t}\left(a e^{-b t}\right) \
&=-a b e^{-b t}
\end{aligned}
$$
The exponential function approaches zero as the exponent gets more and more negative. This means that both the charge and the current are decreasing in magnitude with time. It makes sense that the charge approaches zero, since the balloon is losing its charge. It also makes sense that the current is decreasing in magnitude, since charge cannot flow at the same rate forever without overshooting zero.