Force on a Charge in the Presence of a Conducting Plane

Perhaps the most common procedure in the study of any field of physics is to calculate the forces on various objects. Studying force problems in electrostatics is the foundation for developing a strong understanding of electrodynamics. In this topic an example calculation of the force on a charged particle in a system with another particle and a conducting boundary is presented. The “Method of Images” technique is used to solve for the force on our particle of interest due to the presence of the other objects.

Figure 1 shows the setup for this topic. A particle of charge -3q is located a distance of $4z_0$ above a conducting and grounded plane of infinite extent. Another charge, this one of charge +q, is located a distance of $z_0$ above the same plane. Both charges lie on a line perpendicular with the grounded plane.

A reasonable initial thought may be that since the conducting plane has no net charge it will have no effect on the charged particle located at $4z_0$ . While it is true that there is no net charge in the grounded plane, it can still be affected by the charged particles near it. These charges induce a polarization in the grounded plane and cause its charges to rearrange themselves on the surface. A conceptualization of this effect for each of the charges is shown in figure 2. The free electrons (this is a conducting plane so it has free electrons) in the plane will be attracted to the positive charge at $z_0$ and repelled by the negative charge that is further away. While the negative charge is larger, since it is further away from the plane it may not have as significant an effect as the closer positive charge.

polarization due to charges — Fig. 2: Polarization of the conducting plane due to the charged particles near it.

To determine the electrostatic force on the charge of value -3q we need to determine the final charge density of the grounded plane. This has quickly become a difficult calculation.

Fortunately, in electrodynamics it is possible to design alternative systems that correctly describe the features of our present system. The grounded plane can be replaced with a collection of charges as long as those charges result in the same grounded potential (i.e., zero potential) at the position of the plane.

This is an application of the uniqueness theorem, which basically means that if two electrostatic systems have the same boundary conditions, then they will have the same field structure. Our primary boundary condition in this example is the grounded plane. Figure 3 illustrates the positioning of image charges that replicates the boundary condition of the grounded plane.

Considering the geometry laid out in figure 3, notice that the image charges are placed exactly opposite to the pre-existing charges. This is a conceptual matter: generating a zero potential boundary at the location of the grounded plane is achieved by placing new charges that cancel out the potential contributions of the existing charges. Since potential is additive, the easiest way to cancel out the pre-existing charges is by placing new ones that are mirror reflections in space and opposites in charge. The figure shows this setup (the z-positions are written in square brackets and all of the charges lie on the z-axis).

At this point there are now even more charges than before, so has the difficulty of the calculation been reduced? It has, specifically because we no longer have a conducting plane in the system and therefore it is not necessary to determine the surface charge density induced by the other charges. The force on the particle of charge -3q, $F_{-3q}$ , is given by the sum of the forces exerted on it from each of the three other charges,

$\vec{F}_{-3q} = \vec{F}_{-3q,+q} + \vec{F}_{-3q,-q} + \vec{F}_{-3q,+3q}$

where $\vec{F}_{a,b}$ denotes the charge on particle a due to particle b.

While there are three separate terms in the expression above, each one is simple. The force on a charged particle of charge $q_1$ , due to a particle of charge $q_2$ is given by,

$\vec{F}_{1,2} = q_1\vec{E}_2$

where $\vec{E}_2$ is the electric field of particle $q_2$ at the location of particle $q_1$ .

The final piece of this puzzle is the electric field due to a point charge,

$\vec{E}_2 = \frac{q_2}{4\pi \epsilon_0 r^2} \hat{r}$

where this particle is located at the center of a spherical coordinate system in which $\hat{r}$ is the radial unit vector.

Our system is simplified to a one dimensional problem as laid out in figure 3. The force is contained entirely along the z-axis. The force on the particle of interest is,

$\vec{F}_{-3q} = (-3q) \frac{q}{4\pi\epsilon_0 (3z_0)^2} \hat{z} + (-3q) \frac{-q}{4\pi\epsilon_0 (5z_0)^2} \hat{z} + (-3q) \frac{3q}{4\pi\epsilon_0 (8z_0)^2} \hat{z}$

$= (-3q) \frac{q}{4\pi\epsilon_0z_0^2} \left[ \frac{1}{9} - \frac{1}{25} + \frac{3}{64} \right] \hat{z}$

$= -\frac{3q^2}{4\pi\epsilon_0z_0^2} (0.117986) \hat{z}$

and the numeric value is not as important as the direction, which is downward. The charge is still attracted toward the oppositely charged particle located just beneath it.

Does this result make sense, or should the grounded plane have had a more drastic effect?

Without a grounded plane the charge on the -3q particle would be given by the first term in the first equation above,

$\vec{F}_{-3q,+q} = (-3q) \frac{q}{4\pi\epsilon_0 (3z_0)^2} \hat{z}$

$= \frac{-3q^2}{36\pi\epsilon_0 z_0^2} \hat{z}$

$= \frac{-q^2}{12\pi\epsilon_0z_0^2} \hat{z}$

and this can be compared to our other result by taking a ratio,

$\frac{\vec{F}_{-3q}}{\vec{F}_{-3q,+q}} = \frac{-\frac{3q^2}{4\pi\epsilon_0z_0^2} (0.117986) \hat{z}}{\frac{-q^2}{12\pi\epsilon_0z_0^2} \hat{z}}$

$= \frac{(3/4)0.117986}{(1/12)}$

$1.06187$

which shows that the force on the -3q particle is greater in the presence of the grounded plane than without.

From a consideration of the image charges it can be determined that our system added a net additional attractive force. We have one image charge with negative charge (repulsive to the -3q charge of interest) at a distance of $5z_0$ . The other image charge is positive and located at a distance of $8z_0$ . Comparing the amplitude of these forces we have,

$\vec{F}_{-3q,-q} = \frac{3q^2}{100\pi\epsilon_0z_0^2} \hat{z}$

$\vec{F}_{-3q,+3q} = \frac{9q}{256\pi\epsilon_0z_0^2} \hat{z}$

$\frac{\vec{F}_{-3q,-q}}{\vec{F}_{-3q,+3q}} = 0.85333$

showing that the force on the -3q charge due to the +3q charge is larger than the force due to the -q charge. Since the +3q charge exerts an attractive force, we expect that the overall effect of these image charges will be to increase the force along the $-\hat{z}$ direction.

This final result can be difficult to understand because it seems as though the presence of the grounded and conducting plane should reduce the force on the -3q charge. This is because it might be expected that the +q charge has a much greater effect on the polarization of the plane than the -3q charge. Since electric field falls off as $1/r^2$ and the +q charge is 4 times closer to the plane than the other charge it should have 16 times the effect (“effect” has been equated to electric field at the plane). The -3q charge has 3 times the charge, but that is more than canceled out by the further distance.

Consider the far edges of the grounded plane, however, and a different result comes to mind. Far out along the plane it will seem as though the charges are the same distance away. Figure 4 displays the geometry. As we move further along the y-axis the total separation between the grounded plane and the charges approaches $y_i$ . The total distance, r, is given by,

$r = \sqrt{z^2 + y_i^2}$

but as $y_i >> z$ this becomes approximately $r = y_i$ . At these locations along the grounded plane the charges appear to be the same distance away but one of them has three times greater charge. It is suggested that the larger absolute charge has more influence at large distances. Perhaps the -3q charge results in more positive surface charge buildup at these locations. Such a buildup would then help to explain why the effect of the grounded plane is to increase the $-\hat{z}$ ” force on the charge. The counter argument to this is that as the distance increases too much the effect of the charges goes to zero precisely because the denominator in the electric field expression grows so quickly.

Does this result make sense, or should the grounded plane have had a more drastic effect?

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