When physicists calculate the value of a physical quantity from an equation, they pay particular attention to the units involved. A force of 2 is ill-defined, a force of 2 Newtons is clear. Units tell you both what type and how much.
However, when physicists want to check the plausibility of an equation, without worrying exactly about “how much” and which set of units will be used (e.g. Newtons vs. pounds vs. dynes), they often look at the “dimensions" of the physical quantities involved. Dimensions are the powers of the fundamental physical quantities: length (\(L\)), time (\(T\)), mass (\(M\)), and charge (\(Q\)), that make up the physical quantity. They only tell you what type of quantity and not “how much”. For example, since force is mass times acceleration,the dimensions of force are \(ML/T^2\). A common notation for the dimensions of a quantity is to put the quantity in square brackets, for example:
\[[Force]=\frac{ML}{T^2}\]
Find the dimensions of electrostatic potential energy. Also, find the dimensions of electrostatic potential.
For electrostatic potential energy:
Energy is energy. I like to remember that kinetic energy is \(\frac{1}{2} mv^2\) so the dimensions are \begin{equation*} [U]=M \frac{L^{2}}{T^{2}} \end{equation*}
For electrostatic potential:
Electrostatic potential is the potential energy per unit charge, so I divide the resulting dimensions above by charge. \begin{equation*} [V]=\frac{ML^{2}}{Q\,T^{2}} \end{equation*}
Alternatively, I can remember the iconic equation for the potential due to a point charge, \(V=k\frac{Q}{r}\), but then I need to know the dimensions of \(k\). If you remember these, then great, but I never do. So I compare the iconic equation for the force between two charges, \(\vec{F}=k\frac{q_1q_2}{r^2}\hat{r}\) to the iconic equation for the potential, since both have a factor of \(k\). I recognize that if I find the dimensions for force, I should multiply those dimension by a length and divide by a charge. Force is force, so I get the dimensions from \(\vec{F}=m\vec{a}\). Don't forget the unit vectors are dimensionless!
\begin{align} [\vec{F}]&=M\frac{L}{T^2}\\ \Rightarrow [V]&=\frac{ML}{QT^2} \end{align} These are the same dimensions that I got by the other method.
For each case below, find the total charge.
The charge density depends only on \(r\), and is zero for \(r<a\) or \(r>b\). The total charge is given by \begin{align*} \int\limits_{\hbox{shell}} \rho(\vec{r}) d\tau &= \int\limits_0^{2\pi}\int\limits_0^\pi\int\limits_a^b 3\alpha\, e^{(kr)^3} r^2\sin\theta\,dr\,d\theta\,d\phi\\ &= \frac{4\pi\alpha}{k^3} \, e^{k^3 r^3} \Bigg|_a^b\\ &= \frac{4\pi\alpha}{k^3}\left(e^{k^3 b^3}-e^{k^3 a^3}\right) \end{align*} Note that \([k]=\frac{1}{L}\) so that the argument of the exponential is dimensionless and \(\alpha\) has dimensions of volume charge density, \([\alpha]=\frac{Q}{L^3}\).
The charge density depends only on \(s\), and is zero for \(s<a\) or \(s>b\).
This problem is ill-posed, since the height of the cylinder is not given. Assuming that the cylinder extends from \(z=A\) to \(z=B\), the total charge is given by \begin{align*} \int\limits_{\hbox{shell}} \rho(\vec{r}) d\tau &= \int\limits_A^B\int\limits_0^{2\pi}\int\limits_a^b \alpha\, \frac{1}{s}\, e^{ks} s\,ds\,d\phi\,dz\\ &= \frac{2\pi\alpha}{k} (B-A) \, e^{ks} \Bigg|_a^b\\ &= \frac{2\pi\alpha}{k} (B-A) \, \left(e^{kb} - e^{ka}\right) \end{align*} Note that \([k]=\frac{1}{L}\) so that the argument of the exponential is dimensionless and \(\alpha/s\) has dimensions of volume charge density, so that \([\alpha]=\frac{Q}{L^2}\).
If the cylinder is infinitely long, then the total charge is of course also infinite. But a better answer in this case would be that the charge per unit length is \begin{equation*} \frac{2\pi\alpha}{k} \left(e^{kb} - e^{ka}\right) \end{equation*}