Irodov Problem 3.66

(a) Say, plates 2 an 3 are connected across the terminals of a battery as shown. The battery sucks out electrons from plate 2 making it positively charged and deposits them on plate 3 making it negatively charged. Since the total charge has to be conserved, the total charges on plates 2 and 3 must be equal and opposite. Further, all charge within the plates will be uniformly distributed since electrons repel each other and would like to be as far from each other as possible. So let the charges on plates 2 and 3 be and respectively.

Since plates 1 and 4 are connected through a conducting wire, electrons are also free to flow between the plates. Consequently the total charge across plates 1 and 4 must also be conserved. Further, plates 1 and 4 must be at the same potential since they are connected by a wire. Hence, we have,



Since the potential difference between plates and 2 ans 3 is , thus,



Let us arbitrarily fix



From (2) and (3) we have,



Since, the potentials on plates 1 and 4 are the same, this must imply that and . This means that the direction of electric E12 will be from plate 2 towards plate 1 and that of E34 will be from plate 4 towards plate 3 as shown in the figure.

Now let us put a Gaussian cylinder of area A as shown in Figure 2.



The total electric flux through the cylinder will be . The total charge contained in the Gaussian cylinder is . From Gauss law we have,



The potential at plate 1 will be and that of plate 2 will be . Since both these potentials must be the same (1), we have,






Similarly, since the potential between plates 2 and 3 is we have,





(b) The charge densities of plates 1 and 4 will be equal and opposite as the total charge across them is conserved. Let these charges by and respectively. Let us first find the electric field due to a charged plate with surface charge density .


Let us consider a Gaussian cylinder across the charged conductor with area A. Let the electric field be E due to the charged conductor. The total electric flux through the cylinder will be 2EA. The total charge contained within this cylinder is . Hence, from Gauss Law we have,



For our problem then, the net electric filed is the superposition of the electric fields from all the plates. This is illustrated in Figure 4.



As seen from Figure 4,







Also we have,

Irodov Problem 3.65

Since electrons are free to move about within a conductor, and two electrons repel from each other, they will tend to spread themselves uniformly across the surface of the conductor. Given the spherical symmetry of the problem, the spherical shells will have a uniform charge density. Thus, the surface charge densities of the spheres will be



The electric potential at the center of the sphere O, will be due to each of the two spheres, equal to,








If this potential Vo is zero then from (2) we have,




Consider a spherical Gaussian surface within the innermost sphere (G1 as shown in the figure). Since there is no charge within this sphere, there must be no electric flux through G1, in other words the net electric filed inside the sphere will be zero at all points.

Consider another spherical Gaussian surface (G2 as shown in the figure) with radius r such that . Since the net charge inside this sphere is q1, the electric field in this region can be determined using Gauss law and is given by,




Now consider another spherical Gaussian surface (G3 as shown in the figure) with radius r such that . The net charge within this Gaussian surface is . Hence, from Gauss law, the electric field in this region is given by,



The electric potential V(r) as a function of distance is given by,




Using (4a) and (4b) we have,

Irodov Problem 3.64

First let us start by understanding some important concepts. As described in Problem 3.63, the entire conductor must have the same potential and hence, there cannot be any electric field inside the conductor. In other words there is no electric field within the conducting shell.

Consider a Gaussian spherical surface of radius such that ., in other words this Gaussian surface is just above the inner circle within the conductor. Since there cannot is any electric field in side the conductor, there is no electric flux through the Gaussian surface. From Gauss Law in turn, this imples that the net charge contained within this Gaussian sphere must be 0. This can only happen if there is a net charge of -q on inner the surface of the sphere. Let be the charge density at a location in spherical coordinates on the inner surface of the conductor. Then we have,

Since the electric field anywhere within the conductor is 0, for all Gaussian surfaces in the conducting shell with radius R such that the net flux through them will be zero, in turn implying that no free charges now exist anywhere within the shell. Since the total charge in the conductor must be consrved, this in turn implies that all the positive charges in the shell are concentrated on the outer surface of the conducting shell as shown in the figure. Further, the sum total of all these positive charges will be equal to q. Let be the surface charge density on the outer surface of the conductor at a location with sperical coordinates . Then, we have,

The potential at the center of the sphere is due to three sources, i) the charge q, ii) the surface charge density at the inner surface of the conducting sphere and iii) the surface charge density at the outer surface of the conducting sphere and is given by.