Irodov Problem 3.33

The disc can be regarded as to be composed of several infinitesimally thin rings. Let us consider one such ring of radius r and thickness dr. The charge contained in this ring will be,



Using the expression derived for the electric field due to a ring in Problem 3.9 Eqn 1, the electric field due to the infinitesimally thin ring is give by,



Similarly using the expression derived in Problem 3.30 Eqn 1, we determine the potential as,













When l tends to 0, From (2) and (3),







When l much greater than R, using Taylor's series,






Using (4a) and (4b) respectively on (2) and (3) we obtain,

Irodov Problem 3.32

The hollow hemispherical bowl can be thought of being composed several infinitesimally thin rings. Consider an infinitesimally thin ring (shaded as red in the figure) that is at an elevation angle and its thickness subtends an angle . The thickenss of this ring will be and its radius will be . The total surface area of this thin ring will hence be . The total charge contained in this ring will hence be,

Now in problem 3.9 we have already computed the electric field due to a ring at points along its axis. Since, the center of the hemisphere is at a distance from this infinitesimally thin ring
(see in the figure), using this result we can compute the electric field due to this inifinitesimally thin ring as,

Isnt't it interesting that the electric field strength at the center does not depend on the radius R of the bowl? The net electric field can be determined as,

Similary, we can determine the potential at the center of the infnitesimally thin ring by using the result derived in Problem 3.30 Eqn 1 and use it to determine the potential at the center of the hemisphere as,

Irodov Problem 3.31

From Problem 3.22 Eqn 1 we know that the field due to an infinitely long thread with uniform charge density at a distance r from the thread is given by,




directed along the radial direction in the horizontal place perpendicular to the thread (as shown in the figure).

The fact that electric field decays inversely as distance from the thread makes this problem very interesting. This problem cannot be solved using Method 1 as we did in case Problem 3.30. The reason is that every point in space around the thread has an inifinite electric potential energy. In other words it will require an infinite amount of work to bring any charged particle from infinite distance to anywhere close to the thread. Alternatively said, any charged particle left on its own from any point close to thread will acquire an inifinite amount of kinteic energy by the time it reaches far away. This is a direct consequence of the fact that the integral of 1/x diverges to infinity. However, the potential difference between two locations finitely dustance from the thread will be finite.

Consequently, we shall use Method 2 (see 3.30) to solve this problem. The potential difference between two points at distances at distances r1 and r2 is given by,