1 Crore+ students have signed up on EduRev. Have you? 
Method of Images
The uniqueness theorem for Poission’s or Laplace’s equations, which we studied in the last couple of lectures, has some interesting consequences. Frequently, it is not easy to obtain an analytic solution to either of these equations. Even when it is possible to do so, it may require rigorous mathematical tools. Occasionally, however, one can guess a solution to a problem, by some intuitive method. When this becomes feasible, the uniqueness theorem tells us that the solution must be the one we are looking for.
One such intuitive method is the “method of images” a terminology borrowed from optics. In this lecture, we illustrate this method by some examples.
Consider an infinite, grounded conducting plane occupying which occupies the xy plane. A charge q is located at a distance d from this plane, the location of the charge is taken along the z axis. We are required to obtain an expression for the potential everywhere in the region z > 0 , excepting of course, at the location of the charge itself. Let us look at the potential at the point P which is at a distance r_{1} from the charge q (indicated by a red circle in the figure).
Instead of solving the problem from the first principles, let us suppose there existed an “image charge” q’ at a distance d’ “below” the plane. This is very similar to the image of an object located in front of a mirror, the image of the object is “behind” the mirror. The image is virtual because we cannot capture the image on a screen behind the mirror. In the same way, the image charge is located at a virtual distance and the charge itself is fictitious. Let the point P be at a distance r_{2} from the image charge. We take the origin on the plane, as shown. Let the position of P be (x,y,z).
The potential φ_{1}(P) at P due to the real charge q and φ_{2}(P) due to the image charge q’ is given by superposition,
The potential P is given by Note that this expression for the potential satisfies the Laplace’s equation at a general point P above the conducting plane, since and at a point P, neither of the distances is zero so that the delta function vanishes. Let us impose the boundary condition of the potential being zero on the boundary, We must have, on the plane, Since the distances are positive, the chargesq and q’ must have opposite sign. V Further, we must have, substituting z = 0 for the position of the conducting plane,
Rewriting this as,
This equation must be satisfied at all points on the conducting plane, I.e for arbitrary values of x,y. Thus the two terms of the above equation must be separately zero. This gives us the solution,
Thus the image charge is equal and opposite to the object charge and, like in the case of optical images, the image distance is equal to the object distance.
It is important to realize that the field exists only in the region above the conducting plane. The expression for potential which satisfies Laplace’s equation everywhere above the plane and also satisfies the boundary condition on the plane, is given by
where we have, because of obvious reasons, switched to a cylindrical coordinates as the square of the distance of the foot of the perpendicular from the point P onto the plane from the origin.
The electric field is obtained as the negative gradient of the potential, and is given by
We emphasize that this expression is valid only in the region From this expression for the electric field, we can obtain the charge density induced on the surface. Remembering that charge density is given by the normal component of the electric field, we only need to evaluate the z component of the electric field at z=0,
The total induced charge is obtained by integrating this expression in the entire xy plane,
Thus the total induced charge has the same magnitude as the real charge, though of opposite sign. In the following figure we have plotted the charge density as a function of xy coordinates on the plane for some representative distances of the object charge. One can see that the magnitude of the charge density is maximum at a point on the plane directly opposite to the real charge and decreases as the distance from this point increases.
The following figure shows the lines of force and equipotentials. As expected the lines of force strike the conducting plane normally. The equipotentials are spheres (shown as circles)
Thus in the above problem we have replaced the original problem of a charge and a conducting plane by the charge and an image charge and eliminated the conducting plane altogether. The solution obtained satisfies the boundary condition that at z=0, the potential is zero. The solution does not make sense in the region z < 0 but that is immaterial because we did not seek the solution in that region in any case! The electric field on the plane z = 0is purely along the vertical plane because, the object and the image charges being equidistant from a point on the conducting plane, the horizontal components cancel by symmetry. The uniqueness theorem guarantees that this is the only possible solution.
Let us calculate the field that is generated at the position of the real charge due to the induced charges. For this we take the general expression for the electric field and consider only the contribution due to the image charge, given by the second term within each curly bracket. As the real charge is located at p = 0, z = d, only the normal component gives nonzero contribution,
Thus the force on the charge due to the charges induced on the surface of the conductor is which is the same as the force exerted by the image charge on the real charge (located at a distance 2d from each other.
It is interesting to calculate field at the surface due to the charge q. In this case, we do a bit of hand waving and consider only half of the field that we have calculated. This is because the image charge being fictitious does not create a field on the surface of the conductor. The force on the conductor is then given by
The force exerted on the surface is then
which, as expected from Newton’s third law, is equal and opposite to the force exerted on the charge by the surface.
We can now proceed to calculate the electrostatic energy of the system. This is just the work done in bringing the charge from infinity to its position on the z axis. We already have an expression for the force exerted on the charge when it is at a distance d from the surface. It is a simple matter to get the expression for the force when the charge is at a distance z along the z axis. Since the electrostatic force is conservative, we bring the charge along the z axis. The work done is then
Example 1
Consider two semi infinite grounded, conducting planes intersecting along the z axis, which is taken out of the plane of the paper (which is the xy plane). A charge q is located at a point (a, b) Calculate the force on the charge q due to the charges induced on the surfaces of the planes.
The image charges are also line charges of charge density λ and In addition, there is a line charge of density + λ at The potential due to a line charge parallel to z axis is independent of z coordinate and only depends on the distance p of the point from the line charge, being given by,
where C is a constant and where is the coordinate of the point P where the potential is sought. Similar expressions for the potential due to the image charges can be written down and the total potential obtained by superposition. Thus, the net potential at is given by
It can be seen that the solution satisfies the boundary condition Note that the tangential components on the y = 0 plane is and that on the x = 0 plane are zero, as expected. The normal components, from which we can infer the charge densities are along the positive y direction for the former and along the positive x direction for the latter.
In the following figure we have plotted the charge densities for different locations of the given charged line indicated on the graph. Consider the situation where the line charge is at the point (a = 2, b= 1). The axes are the coordinates of points on the plane.
Note that the charge density is maximum on the plane at the point (2,1) and spreads out when one goes away from this point. If you reduce the x distance to, say, [1,1], the charge density has a smaller peak and spreads out more.
The total linear charge density on the y=0 plane can be calculated by integrating the charge density
The linear charge density on the x=0 plane can be written down by symmetry to be
Adding these and using the identity tan we get the total linear charge density on the two planes to be λ.
The method of images is not restricted to only planar geometry. We will take up different application in the next lecture. (Part of the video for application of method of images applied to spherical conductors appears in this unit but the text for the entire problem appears along with Lecture 18].
Tutorial Assignment
1. A point charge is located at a distance d above a grounded, infinite, conducting plane. Let P be the foot of the perpendicular from the position of the charge on to the plane. Find the radius of a circle centred at P in which one quarter of all the induced charges remain.
2. A point charge q is located at a point P along the bisector of the angle between two semiinfinite grounded conducting planes at a distance d from the intersection of the planes. The angle between the planes is 600. Determine all the image charges. Find the potential at a point S, located at a distance r from the intersection along the line joining P and S.
Solutions to Tutorial assignment
1. We had shown that the charge density is given by
So, the charge induced in a radius R is
2. The image charges along with the charge q forms a regular hexagon of side d, as shown. The image of q at P is –q at P1, the image of –q at P1 is +q at P2, the image of +q at P2 is –q at P3, the image of –q at P3 is +q at P4 and the image of +q at P4 is –q at P5, as shown in the figure.
Note that the distance of P1 and P5 from S is given by triangle law is ( since OP1=OP5=d)
Similarly,
The distances PS = r  d and P3S = R + d.
Using these distances the potential at S can be written as (the signs of the terms are according whether a charge +q or –q appears)
Expand each term in powers of 1/r, keeping terms up to
Self Assessment Quiz
1. A charge q is located z = d at and a second charge 2q at z = 2d above an infinite grounded conducting plane occupying the xy plane at z = 0. Calculate the force exerted on the charge q.
2. A point charge q is located at a distance d above a grounded, infinite, conducting plane. A second charge q is free to move along the perpendicular from the position of the charge q on to the plane. Where should this charge –q be placed so that the net force on it is zero?
Solution to Self Assessment Quiz
1. The charge q and the charge 2q gives rise to two images, the former an image charge located at a distance d below the plane and the latter an image +2q at a distance 2d below the plane. Thus the force (away from the plane) on the charge q is
2. The image charges due to the charges q and q are shown. Let the charge q be at a distance x from the plane.
Force on –q in upward direction is to be equated to zero. This gives
This simplifies to which has the solution
Method of Images for a spherical conductor
The method of images can be applied to the case of a charge in front of a grounded spherical conductor. The method is not as straightforward as the case of plane conductors but works equally well. Consider a charge q kept at a distance a from the centre of the grounded sphere. We wish to obtain an expression for the potential at a point P which is at the position (r, θ), the potential obviously will not depend on the azimuthal angle and hence that coordinate has been suppressed. Let the point P be at a distance from the location of the charge q.
The image charge is located at a distance b from the centre along the line joining the centre to the charge q. The line joining the charges and the centre is taken as the reference line with respect to which the angle θ is measured. Let P be at a distance b from the image charge. Let q’ be the image charge.
The potential at P is given by Using the property of triangle, we can express the potential at P as,
Since the potential vanishes at r = R for all values of θ, the signs of q and q’ must be opposite, and we must have,
In order that this relation may be true for all values of θ, the coefficient of cos θ from both sides of this equation must cancel,
Since q and q’ have opposite sign, this gives,
Substituting this in the θ independent terms above, we get,
which gives
It follows that if the object charge is outside the sphere, the image charge is inside the sphere. Using these, the potential at P is given by
The electric field can be obtained by computing gradient of the potential. It can be easily verified that the tangential component of the electric field The normal component is given by,
The charge density on the surface of the sphere is and is given by
The charge density is opposite to the sign of q since
It is interesting to note that unlike in the case of a conducting plane, the magnitude of the image charge is not equal to that of the object charge but has a reduced value,
Thus the potential can be written as
The following figure shows the variation of charge density on the surface as a function of the angle θ. As expected, when the charge comes closer to the sphere, the charge density peaks around θ=0.
Since the distance between the charge and its image a  b is the force exerted on the charge by the sphere is
the force is proportional to the inverse cube of the distance of the charge from the centre. let α be the distance of q from the surface of the sphere,
Consider the limit of (one has to be careful here as the origin would now have shifted to extreme left; instead, measure distance from the surface of the sphere. The charge q is located at a distance from the surface,
The distance of the image charge so that the distance of the image from the surface Thus we recover the case of a charge in front of an infinite plane.
A Few Special Cases
1. If the sphere, instead of being grounded, is kept at a constant potential φ_{0}, we can solve the problem by putting an additional charge distributed uniformly over its surface.
2. If the sphere was insulated, conducting and has a charge Q, we put a charge over the surface after disconnecting from the ground. The potential at any point then is the sum of the potential due to the image problem and that due to a charge at the centre
Method of Inversion
The problem of finding the image charge for a sphere indicates that there is a symmetry associated with finding the potential. The potential expression has a symmetry about the centre of the sphere, which can be termed as the “centre of inversion” and the potential expression is symmetric if we let It may be observed that if is the potential due to a collection of charges then the potential due to the charges located at the inversion points are related by the relation
Consider a charge q located at due to this charge is given by
The potential at due to charges
Thus,
Example : Conducting sphere in a uniform electric field
To produce the field we put a charge and a charge The field at an arbitrary point P is given by
We have, if the charges are far away from the sphere,
so that
so that the situation represents a constant electric field in the z direction with
The potential due to the pair of charges is E_{0} cos θ.
The potential due to the two image charges (shown in the figure, inside the sphere)
Thus the net potential is given by the sum of the potentials due to the charges +Q and the potential due to the image charges,
as was obtained by earlier treatment
Example :A dipole near aconducting sphere
The dipole is placed at the position has a dipole moment The corresponding image charges are located (the direction of the dipole moment being from the negative charge to the positive charge, the image dipole moment vector is directed as shown in the figure sine the image charge has opposite sign with respect to real charge.) The image charges are as follows :
Corresponding to the charge the image charge is
And, corresponding to the charge the image charge is
The image approximates a dipole with dipole moment
where,
We have,
We have,
The dipole moment of the image is thus calculated as follows :
We can rearrange these to write,
The image charges, being of unequal magnitude, there is some excess charge within the sphere. The excess charge is given by
so that,
Example : A charge q in front of an infinite grounded conducting plane with a hemispherical boss of radius R directly in front of it.
The image charges along with their magnitudes and positions are shown in the figure. Here, make the potential on the plane zero, while the pairs q,q’ and make the potential on the hemisphere vanish. By uniqueness theorem, these four are the appropriate charges to satisfy the required boundary conditions.
If we take an arbitrary point the potential at that point due to the four charge system is given by
The electric field is obtained by taking the gradient of the above. As we are only interested in computing the surface charge density, we will only compute the radial component of above and evaluate it at r=R,
Substituting r=R and combining the first term with the third and the second with the fourth, we get,
The charge density can be integrated over the hemisphere to give the total charge on the hemispherical boss,
Example : Line charge near a conducting cylinder
The potential at an arbitrary point p (r, θ) is given by
Equating the tangential component of electric field on the surface to zero, we have,
Taking one of the terms to the other side and cross multiplying,
As this is valid for all values of θ , we have, on equating coefficient of sin θ and that of sin 2θ to zero,
The second relation gives, The second relation gives,
Tutorial Assignment
1. A charge q is placed in front of a conducting sphere of radius R which is maintained at a constant potential φ_{0}. Obtain an expression for the potential at points outside the sphere.
2. A spherical cavity of radius R is scooped out of a block of conductor which is maintained at zero potential. A point charge is placed inside the cavity at a distance r_{0} from its centre. Obtain the image charges, an expression for the potential inside the cavity as also the induced charge on the cavity wall.
3. A conducting sphere of radius R, containing a charge Q, is kept at a height h above a grounded, infinite plane. Calculate the amount and location of the image charges.
4. Two identical spheres of radius R each, contain charge Q and – Q. The spheres are separated by a distance d. Locate the image charges.
Solution to Tutorial Assignment
We have seen that when the sphere is maintained at zero potential, we can introduce an image charge from the centre of the sphere. If we want the sphere to be maintained at a constant potential, we imagine an additional image charge q” at the centre of the sphere which keeps the potential constant at Equating this to φ_{0}, we take this charge at the centre to be equal to The potential at an arbitrary point located at a distance r from the centre is thus given by
2. Let us put an image charge q’ at a distance r_{0}’ along the line joining the centre of the sphere to the position of the charge q which is taken along the direction a point within the cavity along a direction potential at such an arbitrary point is given by
If the point is chosen on the surface of the cavity, this must give zero value for the potential. Following identical method as shown in the lecture, the image charge is found to be and the location of the image charge is The surface charge density is found to be given by
3. Imagine the charge Q to be at the centre of the sphere. The sphere is at a constant potential. In order to keep the plane at zero potential an image charge the plane, i.e. at a distance 2h from the centre of the sphere. This would disturb the potential on the sphere which has to compensated by an image charge from the centre of the sphere. As this image charge is located at a distance from the plane, the image charge due to this is at a distance below the plane, i.e. at a distance from the centre of the sphere. The image in the sphere is a charge distance from the centre of the sphere. Thus we see that there would be infinite number of images below the plane and inside the sphere. There seems to be a pattern for the magnitude and the location of these images. Inside the sphere, at etc. If we define it seems to suggest This assertion can be easily proved as follows. If true, the nth image is at a distance from the plane. The image of this in the plane is at a distance from the centre of the sphere and has a charge The image in the sphere is from the centre of the sphere. These are, respectively, the charge and the distance
Thus we have
Rearranging,
This is a difference equation which can be solved by assuming a solution of the form, which gives which has the nonzero solution It can be seen that in order that the series converges for all distances h > R, the positive square root is to be taken. The total charge in the sphere is
4. Consider the image of charge +Q on the other sphere. It is a charge at a distance from the centre of right hand sphere O’. The distance of this from O is
Similarly, image of Q in the sphere to the left is in the sphere to the left is
It can be seen that there are infinite number of images with the images on the left being
Self Assessment Quiz
1. Consider a point charge in front of an insulated, charged conducting sphere of radius R. If the sphere is to contain a charge Q, what is the potential outside the sphere?
2. A conducting sphere of radius R has a charge Q. A charge q is located at a distance 3R from the centre of the sphere. Calculate the potential at a distance R/2 from the centre of the sphere along the line joining the centre with the charge q.
3. Two point charges q each are at a distance d from each other. What should be the minimum radius R of a grounded sphere that can be put with its centre at the mid point of the line joining the two charges so that the mutual repulsion between the point charges is compensated? Assume d >> R.
4. Two spheres, each of radius R contain identical charge Q. The spheres are separated by a negligible distance. Locate all image charges.
Solutions to Self Assessment Quiz
1. Initially, assume that the sphere is grounded. If the charge q is at the position respect to the centre of the sphere, an image charge satisfies the required boundary conditions. In order that the sphere has a total charge Q, we now disconnect it from the ground and add amount of charge to the sphere, which will be uniformly distributed over the surface. The potential at given by,
2. The image of q in the sphere is a charge from the centre. Since the sphere is a conductor all parts of it are equipotential. As the overall charge is Q, it can be looked upon as an image charge which cancels the potential due to the charge q outside and a charge at the centre equal to The potential of the sphere due to this is
3. Each of the point charge gives an image charge within the sphere with charge at a distance from the centre towards the charge.
These two image charges, having opposite sign to that of the original charges, attract each of them. The distance of these image charges from either of the charges is The net attraction due to these charges must cancel the repulsive force Thus, we have,
Simplifying, we get,
Since b>> R, as a first approximation we can neglect all terms other than the first term on either sides of this equation, and get
4. The problem is very similar to Problem 4 of the tutorial assignment with the difference that both the charges being +Q, the image charges alternate in sign. Taking d=2R, we have, for image on the left sphere,
etc. The total induced charge is
Method of Images for a spherical conductor
Example :A dipole near aconducting sphere
The dipole is placed at the position has a dipole moment The corresponding image charges are located (the direction of the dipole moment being from the negative charge to the positive charge, the image dipole moment vector is directed as shown in the figure sine the image charge has opposite sign with respect to real charge.) The image charges are as follows :
Corresponding to the charge the image charge is
And, corresponding to the charge the image charge is
The image approximates a dipole with dipole moment
where,
We have,
where
We can rearrange these to write,
The image charges, being of unequal magnitude, there is some excess charge within the sphere. The excess charge is given by
so that,
Example : A charge q in front of an infinite grounded conducting plane with a hemispherical boss of radius R directly in front of it.
The image charges along with their magnitudes and positions are shown in the figure. Here make the potential on the plane zero, while the pairs q,q’ and q.q’ make the potential on the hemisphere vanish. By uniqueness theorem, these four are the appropriate charges to satisfy the required boundary conditions.
If we take an arbitrary point P(r,θ), the potential at that point due to the four charge system is given by
The electric field is obtained by taking the gradient of the above. As we are only interested in computing the surface charge density, we will only compute the radial component of above and evaluate it at r=R,
Substituting r=R and combining the first term with the third and the second with the fourth, we get,
The charge density can be integrated over the hemisphere to give the total charge on the hemispherical boss,
Example : Line charge near a conducting cylinder
The potential at an arbitrary point p(r, θ) is given by
Equating the tangential component of electric field on the surface to zero, we have
Taking one of the terms to the other side and cross multiplying,
As this is valid for all values of θ, we have, on equating coefficient of sinθ and that of sin 2θ to zero,
The second relation gives, and substituting this in the first relation, we get,
CONFORMAL MAPPING
A technique which is very useful, particularly in handling two dimensional problems is the method of Conformal mapping which uses complex variable.
We are familiar with the scalar potential corresponding to the electrostatic field. If the electric field is in two dimensions, say, in the x,y plane, we can think of something like a “vector potential” which has only one component in the third dimension, i.e. along the z direction. We had defined, If we define vector potential by the relation then we have, the components of the electric field are given by,
In terms of the “vector potential” which has only one (z) component,\
Thus, we have,
Those of you who are familiar with the theory of complex variables may recognize these equations, albeit with different notations.
Consider functions of a complex variable z = x + iy. The function z → f(z) = w is an image of z plane onto the complex w plane. We write the real and the imaginary parts of
Consider a simple map given by This implies
To see how points in the z plane transform to those in the w plane, consider for instance, three concentric quartercircles in the first quadrant of the z plane, i.e. the x y plane. We have the radii of the circles a = 1,2,3. We know that if we use polar coordinates, the coordinates are given by
The three quarter circles are mapped on to three concentric semicircles of radii 1, 4 and 9 by the mapping f(z) = z^{2}. This is because,
so that the radius of the circles and the range of polar angle is doubled. In the figure we have drawn two straight lines which gets mapped onto straight lines because if the equation to the straight line in the z plane is y = mx, the image in the uv plane is given by
so that straight lines get mapped onto straight lines. The shaded region in the figure to the left maps to the shaded region in the figure to the right.
We define loosely a “well behaved function” as one where the mapping takes neighbouring points in the z plane to neighbouring points in the wplane. In complex variable theory, we define an “analytic function” as a function which is differentiable in the complex plane. There are some subtle points of difference in the differentiability of a function of a real variable and that of a complex variable.
For a function of a real variable x, the derivative of a function at a point x is given by
A natural extension of this definition to the case of complex variable would be to define the derivative at a point z as,
The important difference between the former case and the present case is the way Δz approaches zero. In case of the real variable x, there was just two ways of approaching the point x, from its left or from its right. The only requirement of differentiability was that the left and the right limits must exist and be the same. (This is what made x not differentiable at x = 0). In case of a complex variable, however, there are infinite ways in which the function can approach the point z. Some of these ways are shown in the figure by arrows.
The derivative is defined if the ratio above has a unique value irrespective of the path along which the limit is approached. If, for instance, the function approaches the limit along the real axis, we would have,
On the other hand, if the limit is approached along the positive imaginary axis, we would have,
Since these two ways of approaching the limit must give the same answer, we have, on equating real and imaginary parts of these two expressions
These two relations are known as the CauchyRiemann conditions and are taken as the test of analyticity of a function at a point in the complex plane. Since we are in two dimensions, by taking partial derivative of the former equation with respect to x and of the latter with respect to y and adding them, we get
and, likewise,
Thus the real and imaginary part of w satisfies Laplace’s equation in two dimension. This is the connection of what we are discussing with the electrostatic potential problem.
We had seen that the pair of potentials (φ, A) satisfies,
Thus we can identify (φ, A) with the pair (u, v) f a complex variable and define a “complex potential”
We then have,
Let us look at the lines of force. Remember that the direction of the lines of force is given by
Cross multiplying we have,
Thus the imaginary part of the complex potential is constant along the lines of force. The constancy of the real part φ corresponds to equipotential curves in two dimensions.
If the mapping is such that it preserves the angles between curves in both magnitude and sense, it is known as conformal mapping. (In the following figure, two curves on the left are mapped by some conformal transformation to the curves on the right).
It is known from the theory of complex variables that mapping by analytic functions are conformal, except at the “critical points” where the derivative of the function vanishes. Thus the mapping f(z) = z^{2} is conformal, excepting at the origin.
Example : Conformal mapping f(z) = z^{2}
As was explained earlier, the circle of radius R are mapped onto circles of radius R^{2} in this map. Let us look at how the lines y= constant or x= constant map.
Let y = 2. We then have and v = 2xy = 2xc. We thus have, i.e. which are parabola indicated by blue curves in the figure. These represent lines of force in the case of complex potential formulation.
Similarly when x= constant=c, we have, which are oppositely directed parabola shown by the red curves. These would represent equipotentials.
In applying the technique of conformal mapping to solving electrostatics, we map the given problem to a coordinate system where it is amenable to an easier solution. After obtaining the solution we map it back to the original coordinate system.
Example : Find the electric field in the region between two infinite conducting cylinders of inner radius a and outer radius b. The potential on the inner cylinder is kept fixed at φ_{1} and the outer cylinder at φ_{2}.
As the cylinders are infinite, the potential or the electric field have no z dependence and the variation is only in the two dimensional (r, θ) plane, where the cross sections are circles. We can do a conformal mapping . Since polar equation to a circle is x = r cos θ, y = r sin θ, in the complex z plane, the equation is We will map this onto the wplane by a conformal transformation. An obvious candidate for this mapping is the log function w = Inz which would take the circles r = a and r = b to straight lines on which u is constant, as
so that the circles get mapped to lines u = In r. Thus the two circles become mapped to two straight lines u = In a and u = In b. As we need the expression for the potential between the two circles, we need to solve for u in the region between the two lines.
Since the potential in the original problem was a function of r alone, in the transformed space the potential between the lines would be function of u alone. Since the potential satisfies Laplace’s equation, we have so that the potential has the form To fix the constants, use the boundary conditions when u = In b.
Substituting these the equation to the potential is found to be
We now revert back to the original coordinate system by realizing that u = In r . Thus the potential is given by,
Tutorial Assignment
1. Consider the conformal mapping At which point(s) the mapping is not conformal? How are circles in z plane mapped onto the wplane?
2. Verify that for a complex function is a harmonic function.
3. A wedge shaped region is bounded by two semiinfinite conducting planes intersecting along the z axis, the angle of the wedge being π/3 . The plane at θ = 0 is maintained at zero potential while the one at θ = π/3 is at a constant potential φ_{0}. Find an expression for the potential inside the wedge. [Hint : Consider conformal transformation in two steps, first z : → w = z^{3} and then
4. (Hard Problem) A long cylinder is split into two halves by a thin insulation. One half of the cylinder is grounded while the other half is maintained at a constant potential φ_{0}. Take the length of the cylinder along the z direction. Obtain an expression for the potential inside the cylinder.
Solutions to Tutorial Assignment
1. which vanishes at Hence the function is analytic everywhere except at these points and the mapping is not conformal at these points.
Use the polar representation
so that The equation satisfied by u and v are
Circles in z plane are of constant r. Thus these are mapped on to ellipses, except near the origin where the above equation is not satisfied.
2. It can be easily checked that
3.
The sequence of transformation is shown in the diagram. The location of points is to point out the sequence in which points are mapped and are not to scale. In the first transformation, writing for the coordinates of the inclined plane z^{3} takes it to it is mapped to the negative real axis. The horizontal section remains on the positive real axis, though the distances are scaled as cube of the original distance. (In general if the angle that the incline makes with the horizonta is π/n , the mapping is byz^{n}.
In the second transformation, we use f = 1/π In w so that the original incline which now is along is mapped to so that the real axis is scaled but now Im f = 1. Thus the original inclined plane is now parallel to the original horizontal plane. Clearly, the variation in the potential can only be along the imaginary f direction. Denote the real and the imaginary parts of f as f_{1} and f_{2} respectively. Since the potential satisfied Laplace’s equation, and, taking into account the fact that the only variation is with respect to f_{2}, we have,
Which has the solution φ = af_{2} + b, where a andb are constants to be determined from boundary conditions. We have, φ = 0 when f_{2} = 0 (lower plate), which gives b= 0. Since φ = φ_{0} when f_{2} = 1 (upper plate), we get,
We have to now retrace our path and carry out a sequence of two inverse transformation to go from f plane to w plane and back to z plane.
Now, that its imaginary part is θ. Thus the potential is given by
4. let the length of the cylinder be perpendicular to the plane of the paper. A cross section in the (p, θ) plane is shown in the figure. The radius of the circle is taken to be unity. The upper half is at a potential φ_{0} while the lower half is grounded.
The successive transformation which convert the original problem to the problem of a parallel plate capacitor is shown in the figure. The first transformation takes all the points on the circle to the imaginary axis of the z plane,
The following table shows the location of corresponding points before and after the transformation.
Point  A  B  C  D  E  F 
θ  0+  π/2  π  π+  3π/2  2π 
cot (θ/2)  ∞  1  0+  0  1  ∞ 
Thus the upper half of the cylinder is mapped to the positive imaginary axis and the lower half to the negative imaginary axis. We can write
Which can be written as , the upper sign for the upper half and the lower sign for the lower half. The next mapping takes w to f where
Thus the upper half becomes a line parallel to the real axis at and the lower half a line at
Clearly the potential can only depend on f_{2}. As the potential satisfies Laplace’s equation in only one variable f_{2}, the solution can be written as where A and B are constants to be determined from the boundary conditions : which gives,
The next task is to express the potential in terms of variables in the w and z plane. We have
where the last step is obtained by writing the imaginary part of the log function by expressing the numerator in a polar representation. Thus the potential is given by
Self assessment Quiz
1. Consider a conformal mapping given by w = sin z. How are lines parallel to x and y axes map?
2. A semicircular region is bounded by the xaxis and a circular arc of radius 1 about the centre. The x – axis is grounded while the circular arc is maintained at a constant potential φ_{0}. Consider a conformal transformation convert the region into a planar region, solve the Laplace’s equation and obtain an expression for the potential in the semicircular region.
3. Two semiinfinite conductors make right angles with each other. The horizontally placed conductor is grounded while the vertical one is at a potential φ_{0}, there being a thin insulation. Find an expression for the potential in the region between the two planes.
Solutions to Self assessment Quiz
1. x = a, we have cosh which are hyperbola with foci at we have so that the equation of trajectory is , which represent ellipses with foci at
2. Consider a mapping For the semicircle of unit radius, so that
with The semicircle is mapped on to the real axis since Thus the semicircular part is mapped along the positive real axis from 0 to infinity.
For the line is purely imaginary and has a value 0 for x=1 and is i^{∞} for x = 1. Thus the real line is mapped along the imaginary axis from the origin to infinity.
For the region bounded by the lines the transformation is given by
which gives,
Since φ, satisfies two dimensional Laplace’s equation subject to boundary condition the solution is given by The potential as a function of (x, y) is obtained by substituting for u and v the expressions above .
3. The solution of this problem proceeds identically with that for the tutorial problem 3 with the change that the first conformal transformation is w = z^{3}. The expression for potential is
11 videos46 docs62 tests

Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 
11 videos46 docs62 tests
