00:01
Hello, in this question we have two rods, one made up of glass and another one which is of plastic.
00:08
And these two rods are charged.
00:11
The glass rod is charged with plus 10 nanoculum while the plastic rod is charged with minus 10 nanoculum.
00:19
Both these rods are separated by a distance of 3 .7 centimetre from one another.
00:26
And the length of the rod mentioned here is 10 cm both the rods have got same length of 10 centimeters each and the distance of separation as i said is 3 .7 centimeter so this is 10 centimeter and also the plastic rod is 10 centimeters in length so obviously charged rods will produce electric field the question asks is what is a electric field at three points, one point, let us say point one is at two one centimeter from the glass rock.
01:07
There's another point two which is at two centimeter from the glass rock and another point three which is at three four three centimeter from the glass rod.
01:17
We need to find out the electric field in between these two glass rod joining this line is joining the midpoints of the glass rods.
01:27
One thing we need to see that one is positively charged while the other one is negatively charged.
01:33
That means the electric seal at any point on this line will be directed in this direction because because of this positively charged glass rod it is in this direction.
01:45
Also because of the negatively charged plastic rod it is also in the same direction.
01:52
So at every point along this line, the electric field is directed towards the right.
01:58
So we need to know how to find out the electric field due to a charged rod of certain length.
02:05
So let us say we have a charge rod of, let us say it's charged up to plus q.
02:11
Then the electric field at a distance x from it and let us say the length of that charge rod is l, then the electric field would be given as k q divided by x under root of x squared plus l square divided by two where it's the length of the rod x is the distance so we are going to use this formula in order to find out the electric field due to both the roads at three different points so let's start doing it so let us say we're talking about the point one first so the electric field same at point 1 because of both the rods.
02:53
So first because of the glass rod it would be k and since the charge for both of them is same and they are bothered about the magnitude let me represent it as q divided by a distance from the glass rod is 1 cm so let's we write it as 10 to the per minus 2 under root of 10 to the power minus 2 square because it's x square plus l squared divided by four plus the length here is 10 cm so we can write it as l divided by two whole square which means 5 multiplied with 10 to the power minus 2 whole square so this is your electric field due to the glass rod similarly the electric field due to the plastic rod same charge so k q divided by distance from here it is 1 centimeter meaning from the plastic rod it will be 2 .7 cm.
03:52
So it is 2 .7 multiplied with 10 to the power minus 2.
03:56
Under root of 2 .2 .2.
03:58
2 .5 multiplied with 10 to the power minus 2 whole square.
04:03
So we have to substitute the value of q here as 10 nanoculum that is 10 multiplied with 10 to the power minus 9.
04:16
And k value is times.
04:18
Value that is 9 multiplied with 10 to the power 9 so we need to substitute all this values and then we will calculate the value for it which will turn out to be 2 .35 multiplied with 10 to the power 5 newton per cooler so the same process goes for point 2 as well as 0 .3 so let us look into point 2 here why from the glass rod the point 2 is that it is 2 cm the point two from the plastic would be at a distance of 1 .7 centimeter.
04:54
So the electric field because of the glass rod would be k, q, the values already mentioned to you, divided by distance which is equal to 2 multiplied with 10 to the par minus 2, under root of 4 multiplied with 10 to the per minus 4, because square of this quantity, plus 25 multiplied with 10 to the pump minus 4...