None of us come to more general charged distribution's.
So in the previous lecture, we have determined the electrostatic potential for a sphere, a four point
charge.
So we have considered that our charge is positioned at the position of zero.
So in this case, our one is equal to zero.
And we found out that the electrostatic potential is just as constant.
Your time times to charge divided by hour.
So in general, if our charge is positioned at some other position, like one in this case, we can
write it down like this.
Now, of course, if we have multiple point charges, we can just add up their individual electrostatic
potentials.
So this is because this equation is including here does the plant's operator.
So that's a sum of second-order partial derivatives.
So the solution can be considered the sum of individual solutions.
So if we have multiple poison charges, we can write down our electrostatic potential just like this.
So, for example, consider we have a charged Q1 at this position or one, and we have another charge,
Q2 at the position or two.
Then we would just have two terms here and here.
We would have Q1 and our one and a plus Q2 and our two.
So what this means is that the electrostatic potential is additive, and if we can write it down as
a sum, we can also write it down as an integral, because maybe you remember this.
If we make our our space smaller and smaller, just some here becomes an integral.
So this means we can write down for a general charged distribution row of R that the electrostatic potential
is just as integral here.
So you see, for in terms of the structure, it looks very much the same as here, it's just that we
have to some here, this is now an integral and here we have expressed our cue in terms of this real
times.
These are so if you remember the charges, just as integral overall times D.R.
So now we have all our tools to consider more difficult examples, and in the next section we will start
by calculating the electric field and the electrostatic potential of an electric dipole.