157 lines
5.6 KiB
TeX
157 lines
5.6 KiB
TeX
\subsection{Electric Field as Force per Unit Charge}
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This subsection defines the electric field from source charges and shows how it determines the force on any charge placed at a point.
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\dfn{Electric field, source charges, and test charges}{Let source charges create an electrostatic interaction in space. Let a field point have position vector $\vec{r}$, and let a small positive test charge $q_0$ be placed at that point. If the electric force on the test charge is $\vec{F}$, then the \emph{electric field} at that point is defined by
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\[
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\vec{E}(\vec{r})=\frac{\vec{F}}{q_0}.
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\]
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The source charges are the charges that produce the field. The test charge is a charge used only to probe the field at a location. Because the factor of $q_0$ divides out, the field depends on the source-charge configuration and the location $\vec{r}$, not on the particular test charge used to measure it. By convention, the direction of $\vec{E}$ is the direction of the force on a positive test charge. The SI units of electric field are $\mathrm{N/C}$.}
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\thm{Point-charge field law and force relation}{Let a point source charge $Q$ be fixed at position vector $\vec{r}_Q$. Let the field point have position vector $\vec{r}$, and define
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\[
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\vec{R}=\vec{r}-\vec{r}_Q,
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\qquad
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R=|\vec{R}|,
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\qquad
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\hat{R}=\frac{\vec{R}}{R},
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\]
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with $\vec{r}\neq \vec{r}_Q$. Then the electric field due to the point charge $Q$ is
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\[
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\vec{E}(\vec{r})=k\frac{Q}{R^2}\hat{R}=k\frac{Q}{R^3}\vec{R},
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\]
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where
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\[
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k=\frac{1}{4\pi\varepsilon_0}=8.99\times 10^9\,\mathrm{N\,m^2/C^2}.
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\]
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If any charge $q$ is placed at that field point, the electric force on that charge is
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\[
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\vec{F}=q\vec{E}.
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\]}
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\pf{Derivation from Coulomb's law}{Let a positive test charge $q_0$ be placed at the field point. Coulomb's law gives the force on the test charge due to the source charge $Q$ as
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\[
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\vec{F}=k\frac{Qq_0}{R^3}\vec{R}.
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\]
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Now divide by $q_0$ and use the definition of electric field:
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\[
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\vec{E}(\vec{r})=\frac{\vec{F}}{q_0}=k\frac{Q}{R^3}\vec{R}=k\frac{Q}{R^2}\hat{R}.
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\]
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This shows that the field is determined entirely by the source charge and geometry. Once $\vec{E}$ is known at a point, the force on any charge $q$ placed there is obtained by multiplying by $q$, so $\vec{F}=q\vec{E}$.}
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\cor{Direction and superposition of electric fields}{Let point source charges $Q_1,Q_2,\dots,Q_N$ be fixed at position vectors $\vec{r}_1,\vec{r}_2,\dots,\vec{r}_N$. Let the field point have position vector $\vec{r}$, with $\vec{r}\neq \vec{r}_i$ for all $i$. Then the net electric field is the vector sum of the individual fields:
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\[
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\vec{E}_{\mathrm{net}}(\vec{r})=\sum_{i=1}^N k\frac{Q_i}{|\vec{r}-\vec{r}_i|^3}(\vec{r}-\vec{r}_i).
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\]
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For a single source charge, the field points radially away from the charge if $Q_i>0$ and radially toward the charge if $Q_i<0$.}
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\qs{Worked AP-style problem}{Two point source charges lie on the $x$-axis. Let
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\[
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q_1=+3.0\,\mu\mathrm{C}
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\qquad\text{at}\qquad
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\vec{r}_1=(-0.20\,\mathrm{m})\hat{\imath},
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\]
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and let
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\[
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q_2=-2.0\,\mu\mathrm{C}
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\qquad\text{at}\qquad
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\vec{r}_2=(+0.30\,\mathrm{m})\hat{\imath}.
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\]
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Let point $P$ be at the origin, so $\vec{r}_P=\vec{0}$. Take $k=8.99\times 10^9\,\mathrm{N\,m^2/C^2}$.
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Find:
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\begin{enumerate}[label=(\alph*)]
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\item the electric field $\vec{E}_1$ at $P$ due to $q_1$,
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\item the electric field $\vec{E}_2$ at $P$ due to $q_2$,
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\item the net electric field $\vec{E}_{\mathrm{net}}$ at $P$, and
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\item the electric force on a charge $q=-4.0\,\mathrm{nC}$ placed at $P$.
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\end{enumerate}}
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\sol First find the displacement vectors from each source charge to the field point $P$.
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For $q_1$,
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\[
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\vec{R}_1=\vec{r}_P-\vec{r}_1=(0.20\,\mathrm{m})\hat{\imath},
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\qquad
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R_1=0.20\,\mathrm{m}.
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\]
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For $q_2$,
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\[
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\vec{R}_2=\vec{r}_P-\vec{r}_2=(-0.30\,\mathrm{m})\hat{\imath},
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\qquad
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R_2=0.30\,\mathrm{m}.
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\]
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For part (a), use the point-charge field law:
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\[
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\vec{E}_1=k\frac{q_1}{R_1^3}\vec{R}_1.
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\]
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Because $q_1$ is positive, the field points away from $q_1$, which at the origin is in the $+\hat{\imath}$ direction. Its magnitude is
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\[
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E_1=k\frac{|q_1|}{R_1^2}=(8.99\times 10^9)\frac{3.0\times 10^{-6}}{(0.20)^2}\,\mathrm{N/C}.
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\]
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So
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\[
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E_1=6.74\times 10^5\,\mathrm{N/C},
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\]
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and therefore
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\[
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\vec{E}_1=(6.74\times 10^5\,\mathrm{N/C})\hat{\imath}.
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\]
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For part (b),
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\[
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\vec{E}_2=k\frac{q_2}{R_2^3}\vec{R}_2.
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\]
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Here $q_2$ is negative, so the field points toward $q_2$. Since $q_2$ is to the right of the origin, the field at the origin is again in the $+\hat{\imath}$ direction. Its magnitude is
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\[
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E_2=k\frac{|q_2|}{R_2^2}=(8.99\times 10^9)\frac{2.0\times 10^{-6}}{(0.30)^2}\,\mathrm{N/C}.
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\]
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Thus,
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\[
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E_2=2.00\times 10^5\,\mathrm{N/C},
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\]
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so
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\[
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\vec{E}_2=(2.00\times 10^5\,\mathrm{N/C})\hat{\imath}.
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\]
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For part (c), add the fields as vectors:
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\[
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\vec{E}_{\mathrm{net}}=\vec{E}_1+\vec{E}_2.
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\]
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Since both fields point in the same direction,
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\[
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\vec{E}_{\mathrm{net}}=(6.74\times 10^5+2.00\times 10^5)\hat{\imath}\,\mathrm{N/C}.
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\]
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Therefore,
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\[
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\vec{E}_{\mathrm{net}}=(8.74\times 10^5\,\mathrm{N/C})\hat{\imath}.
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\]
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For part (d), the force on a charge placed at $P$ is
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\[
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\vec{F}=q\vec{E}_{\mathrm{net}}.
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\]
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Substitute $q=-4.0\times 10^{-9}\,\mathrm{C}$:
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\[
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\vec{F}=(-4.0\times 10^{-9})(8.74\times 10^5)\hat{\imath}\,\mathrm{N}.
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\]
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So
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\[
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\vec{F}=(-3.50\times 10^{-3}\,\mathrm{N})\hat{\imath}.
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\]
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The negative sign means the force points in the $-\hat{\imath}$ direction. Its magnitude is $3.50\times 10^{-3}\,\mathrm{N}$.
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Therefore,
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\[
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\vec{E}_1=(6.74\times 10^5\,\mathrm{N/C})\hat{\imath},
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\qquad
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\vec{E}_2=(2.00\times 10^5\,\mathrm{N/C})\hat{\imath},
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\]
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\[
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\vec{E}_{\mathrm{net}}=(8.74\times 10^5\,\mathrm{N/C})\hat{\imath},
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\qquad
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\vec{F}=(-3.50\times 10^{-3}\,\mathrm{N})\hat{\imath}.
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\]
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