Question

Two semi-infinite grounded conducting planes meet at right angles. In the region between them, there is a point chargeq, situated as shown in Fig. 3.15. Set up the image configuration, and calculate the potential in this region. What charges do you need, and where should they be located? What is the force onq? How much Work did it take to bringqin from infinity? Suppose the planes met at some angle other than; would you still be able to solve the problem by the method of images? If not, for what particular anglesdoesthe method work?

Short answer

The potential is

The force is $F=\frac{q^2}{16\pi {\epsilon }_0}\left\{\left[\frac{a}{{\left(a^2+b^2\right)}^{3/2}}-\frac{1}{a^2}\right]\hat{x}+\left[\frac{b}{{\left(a^2+b^2\right)}^{3/2}}-\frac{1}{b^2}\right]\hat{y}\right\}$, and Workdone to bring q from infinity is $W=\frac{q^2}{16\pi {\epsilon }_0}\left[\frac{1}{\sqrt{a^2+b^2}}-\frac{1}{a}-\frac{1}{b}\right]$, if the planes are placed at arbitrary angles other than 90 degrees, then the problem can't be solved by the method of image charges.

Step-by-step solution

Define an Electric charge.

Electric charge is the fundamental physical property of matter that experiences a force of attraction or repulsion when placed in an electromagnetic field. A unit negative charge can be called an electron, and a positive unit charge is a proton.

Determine the potential in the region.

Two semi-infinite conducting planes are kept at right angles hence from equation 3.9. The potential can be written as follows:

$V(x,y)=\frac{1}{4\pi {\epsilon }_0}\left[\frac{q}{\sqrt{x^2+y^2+{(z-d)}^2}}-\frac{q}{\sqrt{x^2+y^2+{(z+d)}^2}}\right]$

Here, the denominator represents the distance of the x-axis and the y-axis from the point charge +q and –q,

Now, as the point charges and their corresponding images are located on the coordinates of x, y-axis named as a and b, respectively, which means

$x\to (x-a)$and $y\to (y-b)$similarly for image charges it becomes

$x\to (x+a)$and $y\to (y+b)$, therefore putting the values in equation(i) and solve

$$\begin{gathered} V(x,y)=\frac{1}{4\pi {\epsilon }_0}\left\{\frac{q}{\sqrt{{(x-a)}^2+{(y-b)}^2+{\left(z\right)}^2}}+\frac{q}{\sqrt{{(x+a)}^2+{(y+b)}^2+{\left(z\right)}^2}} \\ -\frac{q}{\sqrt{{(x+a)}^2+{(y-b)}^2+z^2}}-\frac{q}{\sqrt{{(x-a)}^2+{(y+b)}^2+z^2}} \\ \right\} \\ V(x,y)=\frac{q}{4\pi {\epsilon }_0}\left\{\frac{1}{\sqrt{{(x-a)}^2+{(y-b)}^2+z^2}}+\frac{1}{\sqrt{{(x+a)}^2+{(y+b)}^2+z^2}} \\ -\frac{1}{\sqrt{{(x+a)}^2+{(y-b)}^2+z^2}}-\frac{1}{\sqrt{{(x-a)}^2+{(y+b)}^2+z^2}} \\ \right\} \end{gathered}$$

The placement of the image charges in the field is shown below:

As shown in the above diagram, we require a +q charge and –q charge and their corresponding locations and locations of their images are also shown in the above figure.

Determine the Force on the charged particle.

Since, two like charges are placed in the vicinity, they will experience a force of repulsion. In contrast, unlikely that charged two will experience a force of attraction. Therefore, Hence from equation 3.12, the force equation can be written as:

${\overrightarrow{F}}_x=-\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{{\left(2d\right)}^2}\hat{x}$

Similarly, force in the y direction is as follows:

${\overrightarrow{F}}_y=-\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{{\left(2d\right)}^2}\hat{y}$

Therefore, the resultant force can be written as:

Here $\cos \theta =a/\sqrt{a^2+b^2}$and,$\sin \theta =b\sqrt{a^2+b^2}$

Therefore, the force equation can be written as:

 Step 4: Determine the Work done (W) to bring the charge from infinity.

Its been known that,

$W=\int F.dl$

Therefore, the Work done to bring the change q from infinity is:

$\begin{array}{rcl}W& =& \frac{q^2}{16\pi {\epsilon }_0}\underset{0}{\overset{\infty }{\int }}\left\{\left[\frac{a}{{\left(a^2+b^2\right)}^{3/2}}-\frac{1}{a^2}\right]\hat{x}+\left[\frac{b}{{\left(a^2+b^2\right)}^{3/2}}-\frac{1}{b^2}\right]\hat{y}\right\}dxdy\\ & =& \frac{1}{16\pi {\epsilon }_0}\left[-\frac{q^2}{2a}+\frac{-q^2}{2b}+\frac{q^2}{\left(2\sqrt{a^2+b^2}\right)}\right]\\ & =& \frac{q^2}{16\pi {\epsilon }_0}\left[-\frac{1}{a}-\frac{1}{b}+\frac{1}{\left(\sqrt{a^2+b^2}\right)}\right]\\ & & \end{array}$

Validate the method of image charges for other angles excluding 90 degrees.

Consider to work,it must be an integer divisor of 180 degrees. Thus 180,90,60 and 45 degrees,the image charge method is used for these angles, but no other angles will work. It works for 45°, say, with the charges as shown in the figure.Consider it works for 45°, then it has been taken that the x-axis is at equipotential, which means thatif the image charge is placed at the reflection point.

Now, by making the 45° line an equipotential, another charge is placed at the image point, which will screw up the x-axis, so another image charge must be entered to reach the equilibrium state. The method of image charges does not work for arbitrary angles. Here, image charges are eventually forced to place within the original region of interest, which is not allowed for all images as they must go outside the region. The diagram is shown below.