The mutual electric force between two charges is given by Coulomb’s law. How to calculate the force on a charge where there are not one but several charges around? Consider a system of n stationary charges $q_1$, $q_2$, $q_3$, ..., $q_n$ in vacuum. What is the force on $q_1$ due to $q_2$, $q_3$, ..., $q_n$? Coulomb’s law is not enough to answer this question. Recall that forces of mechanical origin add according to the parallelogram law of addition. Is the same true for forces of electrostatic origin?
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| Fig 1: A system of three charges |
Experimentally, it is verified that force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to the other charges, taken one at a time. The individual forces are unaffected due to the presence of other charges. This is termed as the principle of superposition.
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| Fig 2: A system of multiple charges |
To better understand the concept, consider a system of three charges $q_1$, $q_2$ and $q_3$, as shown in Fig. 1.8(a). The force on one charge, say $q_1$, due to two other charges $q_2$, $q_3$ can therefore be obtained by performing a vector addition of the forces due to each one of these charges. Thus, if the force on $q_1$ due to $q_2$ is denoted by $\mathbf{F_{12}}$, $\mathbf{F_{12}}$ is given by Eq. (1.3) even though other charges are present.
Thus, $\mathbf{F}_{12}=\frac{1}{4πε_0}\frac{q_1q_2}{r^2_{12}}\hat{\mathbf{r}}_{12}$
In the same way, the force on $q_1$ due to $q_3$, denoted by F13, is given by
$\mathbf{F}_{13}=\frac{1}{4πε_0}\frac{q_1q_3}{r^2_{13}}\hat{\mathbf{r}}_{13}$
which again is the Coulomb force on $q_1$ due to $q_3$, even though other charge $q_2$ is present.
Thus the total force $\mathbf{F_{1}}$ on $q_1$ due to the two charges $q_2$ and $q_3$ is given as
$\mathbf{F_1}= \mathbf{F}_{12} +\mathbf{F}_{13} =\frac{1}{4πε_0}\frac{q_1q_2}{r^2_{12}}\hat{\mathbf{r}}_{12}+\frac{1}{4πε_0}\frac{q_1q_3}{r^2_{13}}\hat{\mathbf{r}}_{13}$
The above calculation of force can be generalised to a system of charges more than three, as shown in Fig. 1.8(b).
The principle of superposition says that in a system of charges $q_1$, $q_2$, ..., $q_n$, the force on $q_1$ due to $q_2$ is the same as given by Coulomb’s law, i.e., it is unaffected by the presence of the other charges $q_3$, $q_4$, ..., $q_n$. The total force $\mathbf{F_{1}}$ on the charge $q_1$, due to all other charges, is then given by the vector sum of the forces $\mathbf{F_{12}}$, $\mathbf{F_{13}}$, ..., $\mathbf{F_{1n}}$:
$\mathbf{F_1}= \mathbf{F}_{12} +\mathbf{F}_{13}+...+\mathbf{F}_{1n} =\frac{1}{4πε_0}\frac{q_1q_2}{r^2_{12}}\hat{\mathbf{r}}_{12}+\frac{1}{4πε_0}\frac{q_1q_3}{r^2_{13}}\hat{\mathbf{r}}_{13}+...+\frac{1}{4πε_0}\frac{q_1q_n}{r^2_{1n}}\hat{\mathbf{r}}_{1n}$
$\mathbf{F_1}=\frac{q_1}{4πε_0}\sum_{i=2}^{n}\frac{q_i}{r^2_{1i}}\hat{\mathbf{r}}_{1i}$
The vector sum is obtained as usual by the parallelogram law of addition of vectors. All of electrostatics is basically a consequence of Coulomb’s law and the superposition principle.
Example 1
Consider three point charges located at the corners of a right triangle, as shown in Figure 3, where:
Find the resultant force exerted on .
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| Fig.3 |
Solution:
Conceptualize: Think about the net force on $q_3$. Because charge q3 is near two other charges, it will experience two electric forces.
Categorize: Because two forces are exerted on charge $q_3$, we categorize this example as a vector addition problem.
Analyze: The directions of the individual forces exerted by $q_1$ and $q_2$ on $q_3$ are shown in Figure 3. The force $\mathbf{F_{23}}$ exerted by $q_2$ on $q_3$ is attractive because $q_2$ and $q_3$ have opposite signs. In the coordinate system shown in Figure 23.7, the attractive force $\mathbf{F_{23}}$ is to the left (in the negative x direction).
The force $\mathbf{F_{13}}$ exerted by $q_1$ on $q_3$ is repulsive because both charges are positive. The repulsive force $\mathbf{F_{13}}$ makes an angle of 45° with the F x axis.
Use Equation $F=\frac{1}{4πε_0}\frac{|q_2||q_3|}{r^2}$ to find the magnitude of $\mathbf{F_{23}}$:
$F_{23}=\frac{1}{4πε_0}\frac{|q_2||q_3|}{a^2}$
$F_{23}=(8.987 \times 10^9 \ Nm^2/C^2)\frac{(2.0 \times 10^{-6} \ C)(5.0 \times 10^{-6} \ C)}{(0.10 \ m)^2}=9.0 \ N$
Find the magnitude of $\mathbf{F_{13}}$:
$F_{13}=\frac{1}{4πε_0}\frac{|q_1||q_3|}{(a\sqrt{2})^2}$
$F_{13}=(8.987 \times 10^9 \ Nm^2/C^2)\frac{(5.0 \times 10^{-6} \ C)(5.0 \times 10^{-6} \ C)}{(0.10 \sqrt{2} \ m)^2}=11 \ N$
Find the x and y components of the force $\mathbf{F_{13}}$:
$F_{13x}=F_{13} \ cos \ 45^0 = 7.9 \ N$
$F_{13y}=F_{13} \ sin\ 45^0 = 7.9 \ N$
Find the components of the resultant force acting on $q_3$:
$F_{3x} = F_{13x} + F_{23x} = 7.9 \ N + (-9.0 \ N) = -1.1 \ N$
$F_{3y} = F_{13y} + F_{23y} = 7.9 \ N + 0 = 7.9 \ N$
Express the resultant force acting on $q_3$ in unit–vector form:
$\mathbf{F_{13}}=(-1.1 \hat{\mathbf{i}}+7.9 \hat{\mathbf{j}}) \ N$
Finalize: The net force on $q_3$ is upward and toward the left in Figure 3. If $q_3$ moves in response to the net force, the distances between $q_3$ and the other charges change, so the net force changes. Therefore, $q_3$ can be modeled as a particle under a net force as long as it is recognized that the force exerted on $q_3$ is not constant.
What if the signs of all three charges were changed to the opposite signs? How would that affect the result for $\mathbf{F_{3}}$?
Answer The charge $q_3$ would still be attracted toward $q_2$ and repelled from $q_1$ with forces of the same magnitude. Therefore, the final result for $\mathbf{F_{3}}$ would be the same.
Example 2
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| Fig.4 |
Conceptualize Because $q_3$ is near two other charges, it experiences two electric forces. Unlike the preceding example, however, the forces lie along the same line in this problem as indicated in Figure 4. Because q3 is negative while $q_1$ and $q_2$ are positive, the forces $\mathbf{F_{13}}$ and $\mathbf{F_{23}}$ are both attractive.
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