00:01
So according to this problem, there are two point charges q1 and q2.
00:07
Q1 has a charge of two microchromes that is 2 times 10 to the minus 6 columns and q2 has a charge of minus 2 microclomes that is minus 2 times 10 to the minus 6 columns and they are placed on the on the two corners of a yeah of a of a square with the side of 3 centimeters.
00:35
That is 0 .03 meters in a side.
00:39
And we have at...
00:41
In two points.
00:45
They've been specified point a, which is at the center of this square configuration.
00:54
And point b, it is at the empty corner closest to point charge q2.
01:02
And we've been also given that the potential due to this point charges far far away is just zero.
01:11
Okay, and we have two tasks to do.
01:13
The first one is we need to calculate the electrical potential at point b.
01:18
Okay, the first thing we do is that we write down the definition of the potential difference.
01:27
This is potential difference and we get to potential difference.
01:36
Immediately so delta v equals minus the the path into all inside the electric field yeah we just yeah the line integral inside an electric field anyway okay now if we take delta v when we see delta this delta always means v final minus v initial meaning this is the potential at the start of the path and this is the potential at the edge at the at the finish of the path anyway even though it seems yeah a little bit cumbersome this whole line integral expression we can simplify it quite easily the first thing we need to understand.
02:39
The first thing we usually do is that we take, in order to define potential, we take the initial far, far away, the initial potential far far away, and we take it as zero.
02:56
Therefore, what we actually calculate is that expression here.
03:01
Okay, we will denote this v final as vr.
03:07
Okay.
03:09
And we're not using vector because even though this is, even though this line integral, actually it's in deep, it's, let me write it this way.
03:20
This is, this is so important.
03:25
Path independent.
03:28
Path independent.
03:30
We can take any path we want.
03:33
Therefore, we're taking a path from infinity, from infinity, far far away from this configuration.
03:40
To the point that it reaches yeah an easy path maybe a straight line that reaches point b okay so all in all we are the potential at point b it's called b m b minus the electric potential door with dr up to the position of our b and we and i've used this another this another thing we can do instead of using dl vector we can use dr and again even though these looks a little bit still looks a little bit compass and we will simplify it quite easily now what is we need to specify the electric field due to the two point charges okay the electric field due to the two point charges, due to a point charge is the following, the typical formula, one of four pi epsilon times the charge of r squared r hat.
05:01
What is this r now? r vector is the relative position vector.
05:12
Okay, and it is always defined as the vector from the charge, two, the observation point, not the other way we were on from the charge to the observation point.
05:23
R of course is the magnitude and our hut is the vector divided by the magnet.
05:29
Okay, now we have two point charges therefore the net electric field at any point will be one of fopi, epsilon knot q1, r1, the relative position vector from charge one to the observation point.
05:49
Let me write it down.
05:52
Let's say this is the observation point.
05:56
Therefore, r1 will be that from q1 to the observation point.
06:01
And r2 will be that one from the charge to the observation point.
06:08
Okay.
06:14
So q1, r1 squared, r1, h2, plus q2, r2 squared, and two, to cut okay the way we combine this these two contributions yeah yeah we did a vector something due to yeah let me write down a superposition principle super position principle which says that in it would be the sum the vector sum for the contributions okay next we need to talk about the potential since we define the credential as v minus the path integral.
07:06
Yeah, from infinity up to rb of the electric field.
07:12
Again, we can split it into 2.
07:16
E1 dotted here minus e2...
