In this lecture we will derive expression for potential at any point on the edge of a thin disk which are density sigma uniformly distributed on the one of the side. It is one of the side that is important. Now to find the potential here phi because of the disk, let us take a arc like this. We will divide the disk in different arcs, let this be the arc and let this be in a circle shape and this be radius of the arc. And we know from the property of the circle, the diameter will make an angle 90 degree when it is joined like this. So this diameter will be twice r, this is twice r and let us take angle theta here and theta will be here. So it is twice theta. What is the length of the arc, the arc length would be equal to r into 2 theta. What is the area of this strip? If we have done this as r in the same direction take d r. So the area of this strip will be this times d r. What would be the charge on this strip d q will be sigma times this. We are sigma is the charge density we know. Now because of this charge, there will be a potential on this point and even if we divide further into sub segments, all the segments are at the same distance r. So the potential would be d phi due to this arc would be equal to the total charge here sigma r twice theta d r upon 4 by E 0 r. This is the, if you integrate it and take theta from here, if we take this will be pi by 2 and this will be 0. Do not take minus pi by 2 to pi by 2 because half of the angle is covering the full segment of arc because we have taken twice theta here. So theta is varying from pi by 2 to 0 r we get cancelled here. What is the relationship between r and theta? We know that cos theta can be written as base upon this because this is 90 degree. So r upon twice r. So d r is equal to minus twice r sin theta d theta put the value of d r here and you get pi by 2 to 0 sigma upon 4 by E 0 keep it separate twice you keep it separate theta you keep it. What is d r? Minus sin a put here twice r sin theta d theta. Now take the constants out. So it is twice twice 4 we will get cancelled it will be sigma r upon pi E 0 integral pi by 2 to 0 minus sin was here theta sin theta d theta. We can integrate this in parts and this will give us minus theta cos theta plus sin theta that is the integral of theta sin theta d theta if you integrate by parts you will get minus theta cos theta like this. Apply the limits pi by 2 to 0 this will become 0 and this will become minus this will become 1. There is a minus sin here since it is pi by 2 to 0 this was becoming minus 1 this will become 0 when you apply the limits. So it will be minus 1 minus sin gets cancelled so sigma r upon pi E 0 is the potential at any point it can be any point on the surface of the disk and you will get the same result the value is pi sigma r upon pi E 0. So that is all thank you.