checkpoint 1

concepts/em/u9/e9-1-electric-potential-energy.tex

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+\subsection{Electric Potential Energy}
+
+This subsection introduces electric potential energy as an energy of a charge configuration and relates it to work done by electric forces.
+
+\dfn{Electric potential energy of a system}{Let a system contain interacting charges. Let $U$ denote the \emph{electric potential energy} of the system, measured in joules. Electric potential energy is a property of the \emph{configuration of the system}, not of one charge by itself. For any two configurations,
+\[
+\Delta U=U_f-U_i,
+\]
+and if the electric force does work $W_{\mathrm{elec}}$ on the system, then
+\[
+W_{\mathrm{elec}}=-\Delta U.
+\]
+Thus, when the electric force does positive work, the system loses electric potential energy, and when an external agent slowly assembles a configuration against the electric force, the system gains electric potential energy.}
+
+\thm{Point-charge pair formula and work relations}{Let two point charges $q_1$ and $q_2$ be separated by distance $r$, and choose the reference value
+\[
+U(\infty)=0.
+\]
+Then the electric potential energy of the two-charge system is
+\[
+U(r)=k\frac{q_1q_2}{r},
+\]
+where
+\[
+k=\frac{1}{4\pi\varepsilon_0}=8.99\times10^9\,\mathrm{N\,m^2/C^2}.
+\]
+If the separation changes from $r_i$ to $r_f$, then
+\[
+\Delta U=U_f-U_i=kq_1q_2\left(\frac{1}{r_f}-\frac{1}{r_i}\right).
+\]
+The work done by the electric force is
+\[
+W_{\mathrm{elec}}=-\Delta U,
+\]
+and for a slow external rearrangement of the charges,
+\[
+W_{\mathrm{ext}}=\Delta U.
+\]}
+
+\nt{Because $U=kq_1q_2/r$, the sign of $U$ depends on the signs of the two charges. If $q_1q_2>0$, then $U>0$ and positive external work is required to bring the like charges closer together. If $q_1q_2<0$, then $U<0$ and the electric force itself tends to pull the unlike charges together. The important viewpoint is that $U$ belongs to the pair of charges as a system. It is not correct to say that a single isolated charge ``has'' electric potential energy by itself.}
+
+\pf{Derivation from quasistatic assembly}{Let charge $q_1$ be fixed, and let charge $q_2$ be brought slowly from infinity to a final separation $r$. Let $x$ denote the instantaneous separation during the motion, with outward radial unit vector $\hat{r}$. The electric force on $q_2$ is
+\[
+\vec{F}_{\mathrm{elec}}=k\frac{q_1q_2}{x^2}\hat{r}.
+\]
+For a quasistatic move, the external force balances the electric force, so
+\[
+\vec{F}_{\mathrm{ext}}=-\vec{F}_{\mathrm{elec}}.
+\]
+The external work done in assembling the pair is the increase in potential energy:
+\[
+U(r)-U(\infty)=W_{\mathrm{ext}}=\int_{\infty}^{r}\vec{F}_{\mathrm{ext}}\cdot d\vec{r}.
+\]
+Since $d\vec{r}=\hat{r}\,dx$,
+\[
+U(r)-0=-\int_{\infty}^{r}k\frac{q_1q_2}{x^2}\,dx
+=-kq_1q_2\left[-\frac{1}{x}\right]_{\infty}^{r}
+=k\frac{q_1q_2}{r}.
+\]
+This also gives
+\[
+\Delta U=kq_1q_2\left(\frac{1}{r_f}-\frac{1}{r_i}\right),
+\]
+and because electric potential energy is defined for a conservative force, the electric-force work satisfies $W_{\mathrm{elec}}=-\Delta U$.}
+
+\qs{Worked AP-style problem}{Two point charges form a system. Let
+\[
+q_1=+2.0\,\mu\mathrm{C},
+\qquad
+q_2=-3.0\,\mu\mathrm{C}.
+\]
+Initially the charges are separated by
+\[
+r_i=0.50\,\mathrm{m},
+\]
+and they are moved slowly until their final separation is
+\[
+r_f=0.20\,\mathrm{m}.
+\]
+Take
+\[
+k=8.99\times10^9\,\mathrm{N\,m^2/C^2}.
+\]
+Find:
+\begin{enumerate}[label=(\alph*)]
+\item the initial electric potential energy $U_i$,
+\item the final electric potential energy $U_f$,
+\item the change in electric potential energy $\Delta U$, and
+\item the work done by the external agent and by the electric force during the slow move.
+\end{enumerate}}
+
+\sol Use
+\[
+U=k\frac{q_1q_2}{r}.
+\]
+Because the charges have opposite signs, the product $q_1q_2$ is negative:
+\[
+q_1q_2=(2.0\times10^{-6}\,\mathrm{C})(-3.0\times10^{-6}\,\mathrm{C})=-6.0\times10^{-12}\,\mathrm{C}^2.
+\]
+
+For part (a), the initial potential energy is
+\[
+U_i=k\frac{q_1q_2}{r_i}
+=(8.99\times10^9)\frac{-6.0\times10^{-12}}{0.50}\,\mathrm{J}.
+\]
+So
+\[
+U_i=-1.08\times10^{-1}\,\mathrm{J}=-0.108\,\mathrm{J}.
+\]
+
+For part (b), the final potential energy is
+\[
+U_f=k\frac{q_1q_2}{r_f}
+=(8.99\times10^9)\frac{-6.0\times10^{-12}}{0.20}\,\mathrm{J}.
+\]
+Thus,
+\[
+U_f=-2.70\times10^{-1}\,\mathrm{J}=-0.270\,\mathrm{J}.
+\]
+
+For part (c),
+\[
+\Delta U=U_f-U_i=(-0.270)-(-0.108)\,\mathrm{J}.
+\]
+Therefore,
+\[
+\Delta U=-0.162\,\mathrm{J}.
+\]
+
+For part (d), because the move is slow,
+\[
+W_{\mathrm{ext}}=\Delta U=-0.162\,\mathrm{J}.
+\]
+The negative sign means the external agent removes energy from the system rather than supplying it. The electric force does the opposite amount of work:
+\[
+W_{\mathrm{elec}}=-\Delta U=+0.162\,\mathrm{J}.
+\]
+
+This result makes physical sense. Unlike charges attract, so when they are brought closer together, the system energy becomes more negative.
+
+Therefore,
+\[
+U_i=-0.108\,\mathrm{J},
+\qquad
+U_f=-0.270\,\mathrm{J},
+\]
+\[
+\Delta U=-0.162\,\mathrm{J},
+\qquad
+W_{\mathrm{ext}}=-0.162\,\mathrm{J},
+\qquad
+W_{\mathrm{elec}}=+0.162\,\mathrm{J}.
+\]