In this **physics video for iit jee**, we will be discussing the topic **electrostatic** capacitance. The topics discussed in this video are,

**Capacitance**- Capacitance of an isolated conductor
- Methods to find capacitance
- The capacitance of a spherical conductor.
- Energy stored in a charged conductor.
- Redistribution of charge.

We will begin with understanding what a **capacitor i**s. The capacitor is nothing but a device that stores charge and potential energy. Capacitance is the measure of the capacitor. In the above video, we have discussed the structure of the capacitor. Capacitors are also known as a condenser. The energy is released slowly or fast depending on the application of the capacitance. In the above video, we have given a diagrammatic description to understand capacitance well.

Mathematically, C =Q/h

Next we discussed the capacitance for an isolated conductor. Suppose we have a conductor and a charge q is provide to it. The voltage ‘V’’ increases. Thus,

c = Q/V

The unit of capacitance is Faraday or ‘F’. The smaller units of farads are,

μF= 10^{-6} F

pF= 10^{-12} F

Now we need a device which stores charge. But there is a limit to which you can provide charge because when we provide charge the potential keeps on increasing. If the potential keeps on increasing then the device will break down. Thus there is a limit up to which we can provide charge to it. If the limit is crossed then, the device breaks down and then charge flows out of the medium. This leads to zero charges in the device.

**The factors affecting capacitance are,**

- The presence of other conductors near the charged conductor.
- Keeping a grounded conductor near to the charged conductor.

Further, we have discussed methods to find capacitance. Suppose take any device, be it a spherical capacitor, cylindrical capacitor, etc. in order to find the capacitance, we will introduce charge ‘q’ into the conducting plates or conductors. Then we will we find the potential difference “V”. The capacity will be C =Q/V. Thus, in the end, we get to know that capacitance is only based on the structure of the device.

Then we have discussed capacitance of a spherical conductor. The radius of the sphere is “R”. There is a charge ‘’q” present on the surface of the sphere. Then the potential V will be,

V =q/(4π∈0R) also We know that, C = q/V

Substituting the value of V in the above equation, we get C = 4π∈0R

From this **physics video for jee**, we understand that that the capacitance depends upon the geometry and not the charge on it.

Next we have discussed energy stored in a charged conductor. Since it is charges, it will definitely have potential energy. If the charge is “Q” then, using equation above.

And now slowly add charge “dq” to it, and then the external work done is,

Then we discussed redistribution of charge. If there are two spheres of radius R_{1 }and R_{2} and have charges q_{1} and q_{2}. Now these two spheres are connected using a conducting wire. Then the potential will be, V1 and V2 ; and if V_{1} = V_{2} then,

If the potential difference is zero, then no charge will flow.

If V_{1} and V_{2 } are not equal to each other then redistribution of charges will take place. Then, P.D. = ∆V ≠0

If we take similar spheres with charges q_{1}’ and q_{2}’ then, q_{1 }+ q_{2} = q_{1}’ +q_{2}’

Finally the potential will be equal to, V_{1}’ =V_{2}’= V and q1’/R1 =q2’/R2

In the CGS system, capacitance is equal to radius. And thus its unit will be cm.

Further in physic video lecture for jee mains we discussed loss of energy during redistribution of charge. The potential energy of a system is less in a spontaneous reaction. The total potential energy is minimum in any system.

∆V= (C1 C2)/2(C1+C2 ) (V1-V2 )^2