Objective :
To find the capacitances of Spherical and Cylindrical capacitors.

Spherical capacitor

A spherical capacitor consists of a spherical conductor of radius ‘a’, surrounded by another hollow concentric sphere of radius ‘b’.

Suppose charge ‘+Q’ is given to the inner sphere and ‘–Q’ to the outer sphere.


We know that field due to ‘charge on outer sphere’ at an interior point is zero.

Hence, field at a point that lies between the conductors depends only on the charge on the inner sphere.

Thus, electric field between conductors at a distance ‘r’ from the center,

E=Q4πε0r2.


Since charge distribution has radial symmetry, potential difference between two conductors is,

V=Va-Vb

V=-∫abE·dr

Upon integration and substitution of limits, we get potential difference between the conductors,

V=Q(b-a)4πε0ba


Capacitance of spherical conductor,

C=QV

Thus, Capacitance of a spherical conductor comes out to be,

C=4πε0ba(b-a).

Cylindrical Capacitor :

Consider a cylindrical capacitor of length ‘L’ ; Radius of the inner cylindrical surface be ‘a’ and that of outer surface be ‘b’.

Suppose a positive charge ‘+q’ is placed on the inner cylinder and a charge ‘–q’ is placed on the outer cylinder.

If the cylinder is long, as compared to the separation between its surfaces, electric field at a point between the cylinders will be radial and its magnitude will depend only on the distance from the axis.

Consider a point ‘P’ between the cylinders at a distance ‘r’ from the axis.

To calculate the electric field at point ‘P’, let us draw a gaussian cylinder of length ‘x’ and radius ‘r’.

The flux through the bases of gaussian cylinders is ‘zero’, as electric field is perpendicular to the base.

Flux through the curved part,

ϕ=∮E·ds

=E(2πrx).

The charge enclosed by the gaussian surface,

Qen=Qlx

Thus, from gauss’s law,

E(2πrx)=Qlx/ε0

Which gives

E=Q2πε0rl.

The potential difference between the cylinders is equal to,

V=-∫baE·dr.

Upon solving the equation, we get,

V=Q2πε0llnba.

The capacitance,

C=QV

Which comes out to be,

C=2πε0lln(b/a).

Summary :
Capacitance of a spherical capacitor with radii ‘a’ and ‘b’ is equal to,

4πε0ba(b-a).

Capacitance of a cylinder capacitor of length ‘L’ with radii ‘a’ and ‘b’, is equal to

2πε0lln(b/a).

Study of Stomatal Distribution on Leaves :-
Stomata are minute pores of eliptical shape, consists of two specialized epidermal cell called guard cells. The guard cells are kidney shape in dicotyledon and dumbell shape in monocotyledon. Stomata control the movement of gases in and out of a leaf, making carbon dioxide available for photosynthesis, and controlling the loss of water from the leaf through transpiration.

Distribution of stomata varies between monocots and dicots, between plant species, and between the underside and top side of the leaves on a plant. Stomata are found more on plant surfaces thriving under higher light, lower atmospheric carbon dioxide concentrations and in moist environments.Usually the lower surface of a dicot leaf has a greater number of stomata while in a monocot leaf they are more or less equal on both surfaces. In most of the floating plants, stomata are found only on the upper epidermis.

This video explains how to study the stomatal distribution on the upper and lower leaf surfaces and to calculate the stomatal index.

objective to find out the emf of an unknown battery using potentiometer let us look at how a potentiometer is used to measure the emf of a battery we know that balancing equation for a potentiometer is b equals to v naught l by l naught where v is the potential difference across the terminals of the potentiometer l is the balancing length v naught is the potential drop across a b and l naught is the length of the wire a b

suppose the terminals of the potentiometer are connected across a battery with emf e at null deflection point the current through the battery is zero and the voltage difference across lm is equal to emf of the battery by identifying the corresponding length and substituting it in the balancing equation the emf of the cell can be measured accurately

one easy way of finding the emf of an unknown battery is to compare it with a standard battery of known emf two batteries one of known emf e1 and another of unknown emf e2 are connected to potentiometer using a two-way key as shown the driving circuit should have a strong battery so that the potential difference across a b is larger than the emf of the two batteries when the key is plugged in s1 the first battery gets connected to the driving circuit when the key is taken out and pressed into the other plug s2 the first battery gets disconnected and the second battery gets connected when the key is plugged into s1 the jockey is slid to find the null deflection point p1 on the wire a b let the balancing length ap1 be l1 e1 is equal to v naught into l1 by l naught where l naught is the length of a b and v naught is the potential difference across it key is taken out and gets inserted into s2 let null deflection occur at a point p2 this time let the balancing length ap2 be l2 e2 is equal to v naught into l2 by l naught upon dividing the two equations we get e2 by e1 equals to l2 by l1 the unknown emf e2 is equal to e1 into l2 by l1

for example in a potentiometer arrangement a battery of emf e1 equal to 5 volts gives a balance point at l1 equal to 240 centimeters now when an unknown battery is connected the balance point shifts to l2 equal to 360 centimeters then the unknown emf e2 is given by the relation e2 is equal to e1 into l2 by l1 thus e2 equals to 5 into 360 by 240 which comes out to be 7.5 volts

summary emf of an unknown battery e2 is equal to emf of a standard battery e1 into balancing length of unknown battery l2 by balancing length of the known battery

Adrenaline 3d Animation