Consider a system of charges $q_1$, $q_2$,..., $q_n$ with position vectors $\mathbf{r_1}$, $\mathbf{r_2}$,...,$\mathbf{r_n}$ relative to some origin (Fig.1). The potential $V_1$ at P due to the charge $q_1$ is
$V_1=\frac{1}{4\pi \epsilon_0}\frac{q_1}{r_{1P}}$
where $r_{1P}$ is the distance between $q_1$ and P.
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| Fig.1: Potential at a point due to a system of charges is the sum of potentials due to individual charges. |
Similarly, the potential $V_2$ at P due to $q_2$ and $V_3$ due to $q_3$ are given by
$V_1=\frac{1}{4\pi \epsilon_0}\frac{q_2}{r_{2P}}$; $V_3=\frac{1}{4\pi \epsilon_0}\frac{q_3}{r_{3P}}$
where $r_{2P}$ and $r_{3P}$ are the distances of P from charges $q_2$ and $q_3$, respectively; and so on for the potential due to other charges. By the superposition principle, the potential V at P due to the total charge configuration is the algebraic sum of the potentials due to the individual charges
$V=V_1+V_2+V_3+...+V_n$
$V=\frac{1}{4\pi \epsilon_0}\left(\frac{q_1}{r_{1P}}+\frac{q_2}{r_{2P}}+\frac{q_3}{r_{3P}}\right)$ (1)
Example 1
An electric dipole consists of point charges $q_1=+12 \ nC$ and $q_2= -12 \ nC$ placed 10.0 cm apart (Fig. 2). Compute the electric potentials at points a, b and c.
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| Fig.2 |
Solution:
Here our target variable is the electric potential at three points, which we find by doing the algebraic sum in Eq. (1).
At point a we have $r_1=0.060 \ m$ and $r_2=0.040 \ m$ so Eq. (1) becomes
$V_a=\frac{1}{4\pi \epsilon_0}\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}\right)$
$V_a=(9.0 \times 10^9 \ N.m^2/C^2)\left(\frac{12 \times 10^{-9} \ C}{0.060 \ m}+\frac{-12 \times 10^{-9} \ C}{0.040 \ m}\right)$
$=1800 \ N.m/C+(-2700 \ N.m/C)$
$=-900 \ V$
In a similar way you can show that the potential at point b (where $r_1=0.040 \ m$ and $r_2=0.140 \ m$) is $V_b=1930 \ V$ and that the potential at point c (where $r_1=r_2=0.130 \ m$) is $V_c=0$.
Let’s confirm that these results make sense. Point a is closer to the 12 nC charge than to the -12 nC charge, so the potential at a is negative. The potential is positive at point b, which is closer to the 12 nC charge than the -12 nC charge. Finally, point c is equidistant from the 12 nC charge and the -12 nC charge, so the potential there is zero. (The potential is also equal to zero at a point infinitely far from both charges.)
Example 2
Two charges $3 \times 10^{–8} \ C$ and $–2 \times 10^{–8} \ C$ are located 15 cm apart. At what point on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
Let us take the origin O at the location of the positive charge. The line joining the two charges is taken to be the x-axis; the negative charge is taken to be on the right side of the origin (Fig. 3).
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| Fig.3 |
which gives x = 9 cm.
If x lies on the extended line OA, the required condition is
which gives x = 45 cm.
Thus, electric potential is zero at 9 cm and 45 cm away from the positive charge on the side of the negative charge. Note that the formula for potential used in the calculation required choosing potential to be zero at infinity.
Example 3
In Fig. 4 a dust particle with mass $m = 5.0 \times 10^{-9} \ kg=5.0 \mu g$ and charge $q_0=2.0 \ nC$ starts from rest and moves in a straight line from point a to point b.
Calculate:
(a) Determine the electric potential at points a and b.
(b) What is its speed v at point b?
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| Fig.4 |
Solution:
(a) We calculate the potentials using $V=\frac{1}{4\pi \epsilon_0}\Sigma_\frac{q_i}{r_i}$
$V_a=(9.0 \times 10^9 \ N.m^2/C^2)\left(\frac{3.0 \times 10^{-9} \ C}{0.010 \ m}+\frac{-3.0 \times 10^{-9} \ C}{0.020 \ m}\right)$
$V_a=1350 \ V$ and
$V_b=(9.0 \times 10^9 \ N.m^2/C^2)\left(\frac{3.0 \times 10^{-9} \ C}{0.020 \ m}+\frac{-3.0 \times 10^{-9} \ C}{0.010 \ m}\right)$
$V_a=-1350 \ V$
(b) Only the conservative electric force acts on the particle, so mechanical energy is conserved:
$K_a+U_a = K_b + U_b$
We get the potential energies U from the corresponding potentials using Eq. (23.12):
$U_a=q_0V_a$ and $U_b=q_0V_b$
We have $K_a=0$ and $K_b=\frac{1}{2}mv^2$. We substitute these and our expressions for and into the energy-conservation equation, then solve for v. We find
$0+q_0V_a=\frac{1}{2}mv^2+q_0V_b$
$v=\sqrt{\frac{2q_0(V_a-V_b)}{m}}$
$v=\sqrt{\frac{2(2.0 \times 10^{-9} \ C)(1350 \ V-(-1350 \ V))}{5.0 \times 10^{-9} \ C}}$
$v=46 \ m/s$
Example 3
A charge $q_1 = 2.00 \mu C$ is located at the origin, and a charge $q_2 = -6.00 \mu C$ is located at (0, 3.00) m, as shown in Figure 5a.
(a) Find the total electric potential due to these charges at the point P, whose coordinates are (4.00, 0) m.
(b) Find the change in potential energy of the system of two charges plus a charge q3 # 3.00 +C as the latter charge moves from infinity to point P (Fig. 5b).
(c) What If? You are working through this example with a classmate and she says, “Wait a minute! In part (B), we ignored the potential energy associated with the pair of charges $q_1$ and $q_2$!’’ How would you respond?
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| Fig.5 |
Solution:
(a) For two charges, the sum in Equation 25.12 gives
$V_P=\frac{1}{4\pi \epsilon_0}\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}\right)$
$V_P=(9.0 \times 10^9 \ N.m^2/C^2)\left(\frac{2.00 \times 10^{-6} \ C}{4.00 \ m}+\frac{-6 \times 10^{-6} \ C}{5.00 \ m}\right)$
$V_P=-6.29 \times 10^3 \ V =-6.29 \ kV$
(b) When the charge $q_3$ is at infinity, let us define $U_i = 0$ for the system, and when the charge is at P, $U_f = q_3V_P$; therefore,
$\Delta U=q_3V_P-0=(3.00 \times 10^{-6} \ C)(-6.29 \times 10^3 \ V)$
$\Delta U =-1.89 \times 10^{-2} \ J$
Therefore, because the potential energy of the system has decreased, positive work would have to be done by an external agent to remove the charge from point P back to infinity.
(c) Given the statement of the problem, it is not necessary to include this potential energy, because part (b) asks for the change in potential energy of the system as $q_3$ is brought in from infinity. Because the configuration of charges $q_1$ and $q_2$ does not change in the process, there is no $\Delta U$ associated with these charges. However, if part (B) had asked to find the change in potential energy when all three charges start out infinitely far apart and are then brought to the positions in Figure 5b, we would need to calculate the change as follows,
$U=\frac{1}{4\pi \epsilon_0}\left(\frac{q_1q_2}{r_{12}}+\frac{q_1q_3}{r_{13}}+\frac{q_2q_3}{r_{23}}\right)$
$U=(9.0 \times 10^9 \ N.m^2/C^2)\left(\frac{(2.00 \times 10^{-6} \ C)(-6.00 \times 10^{-6} \ C)}{3.00 \ m}\right)$
$+(9.0 \times 10^9 \ N.m^2/C^2)\left(\frac{(2.00 \times 10^{-6} \ C)(3 \times 10^{-6} \ C}{4.00 \ m}\right)$
$+(9.0 \times 10^9 \ N.m^2/C^2)\left(\frac{(3.00 \times 10^{-6} \ C)(-6 \times 10^{-6} \ C}{5.00 \ m}\right)$
$U=-5.48 \times 10^{-2} \ J$
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