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**Example 5- Electric field of a finite length rod along its bisector**

Now, instead of dealing with an infinitely long, straight, charged rod, let’s consider a finite length rod and try to figure out its electric field along its bisector.

As another example here, therefore, we can say **E** field of a finite length, uniformly charged rod with length *L* and charge *Q* along its bisector *a* distance away from the rod. In this case, the charged rod that we’re interested in is finite length. Again, the charge is distributed uniformly along the rod to a value of *Q* coulombs, and we’re interested in the electric field it generates along its bisector. Some *a* distance away from the rod, so here is our point of interest.

This example is very similar to the infinite rod example. If we follow similar type of procedure by choosing an incremental element charge element over here and then an arbitrary location within the distribution, which will generate its own electric field at the point of interest, in radially outward direction, since the charge is positive, and since we’re dealing along the points along the bisector, then, we can call this point as our origin, and found a symmetrical one below the origin, symmetrical *dq*, which will in turn generate its own electric field, at the point of interest, and which will be radially outward direction.

Here, also in the case of infinite rods, we will see that when we introduce a coordinate system here, an *x*–*y* coordinate system, such that the point of interest, *P*, is at the origin, resolving these electric field vectors into their components by taking their projections along *x* and *y*, the *y* components will cancel due to their symmetry for being in opposite directions in equal magnitudes.

Since these *dq*’s have the same magnitude and same distance away from the point of interest, therefore the electric fields that they will generate will have the same magnitude so, are their components. As a result of this, the *y* components for aligning in opposite directions with equal magnitudes will cancel, and will be result in electric field, in this case, also going to be in *x* direction only, if we add all these *dEx*’s to one another, the *x* components, by taking their integral, we’re going to end up with the total electric field along *x*-axis.

Now, up to this point analysis is exact the same with the infinite rod problem, and so I’m not going to go into the detailed calculations once more. We will just go back and look at our integral that we set up for the infinite case. The integral eventually turns out to be, something like this over here. In this case, we can write down the linear charge density in explicit form because we know the length of the rod as well as the amount of charge that we have along the distribution.

The only difference is when we add all the associated *dq*’s one another, our boundaries of the integration will go along the finite, length of the rod. So it will from minus infinity plus infinity. If we choose our origin over here, the boundaries of the integration, *y*-coordinate, is going to vary from this end to the upper end along the length. And if this is origin, this will be the half of the rod, so relative to do this origin, this end will be minus *L* over 2 and this end will plus *L* over 2. Therefore, we can a note over here by saying that integral will be identical to the infinite rod example except the boundaries of the integration.

Then, the electric field is going to be equal to, if we go back to that original equation over here from the infinite rod, we will have *λR* over 4*πε*0, *λR* over 4*πε*0, and inside of the integral, we will have *dy* over *y*2 plus *R*2, integral of *dy* over *y*2 plus *R*2 to the power 3 over 2. And the *y* vary from minus *L* over 2 to plus *L* over 2.

Again, the integration will be exactly identical to the previous case. We will use the similar substitution, we will say let *y* is equal to *R* tangent 5, so on and so forth, and eventually it will give us *λR* over 4*πε*0 cosine *Φ dΦ*. *λR* over 4*πε*0 integral of cosine *Φ dΦ* integrated from *Φ*1 to *Φ*2, which will give us . . . oh, here we have to be careful, maybe I should go a little bit more back and write these in explicit form.

Since *y* is equal to *R* tangent *θ*, *dy* is going to be equal *R* over cosine squared *Φ dΦ *and the denominator which is *R*2 plus *y*2 to the power 3 over 2, is going to give us, as in the infinite rod example, *R*3 over cosine cubed *Φ*.

Substituting those into our integrand, we will have *R* over cosine squared *Φ*,* dΦ* divided by *R*3 over cosine cubed *Φ*, evaluated at *Φ*1 to *Φ*2. Again, this *R* and that *R* will make *R*2 which will cancel the *R*3, leaving *R* in the denominator. We will have electric field is equal to *λ* over 4*πε*0*R* integral of cosine *Φ* *dΦ*, evaluated at *Φ*1 and *Φ*2. And going back to the original variable, electric field is going to be equal to *λ* over 4*πε*0*R*.

Let’s take this integral, which will give us sine *Φ*, which be evaluated at *Φ*1 and *Φ*2. Now going back to the original variable, and applying the exact the same procedure for the previous example, electric field will be *λ* over 4*πε*0*R* and sine *Φ* was, from the previous example, *y* over *y*2 plus *R*2, the square root, which is going to be evaluated now at minus *L* over 2 and plus *L* over 2 in terms of the original boundaries, relative to the origin of variable.

Therefore, electric field is going to be equal to, *λ* over 4*πε*0*R*. First we’re going to substitute *L* over 2 for *y*. So in the numerator we will have *L* over 2 divided by *R*2 plus square of *L* over 2 is *L*2 over 4, minus, we will substitute –*L* over 2 now for *y*, so –*L* over 2 divided by square root of *R*2 plus square of –*L* over 2 is going to give us, again, *L*2 over 4.

As we can easily see, this minus times that minus will give us plus, and denominator is the same so if we add *L* over 2 to *L* over 2, we’re going to end up with *L*. Electric field, therefore will be equal to *λ* over 4*πε*0*R* times *L* over, let’s have a common denominator here. It’s going to give us 4*R*2 plus *L*2 divided by 4 in square root.

Now, *λ* is the total charge of the distribution, which is *Q* divided by total length of the distribution, and that is *L*. Substituting that for *λ*, we will have *Q* over *L* divided by 4*πε*0*R*, and we have here times *L* divided by 4 is going to come out from the square root as 2, so we will have 1 over 2 times 4*R*2 plus *L*2 squared.

And here, this *L* and that *L* will cancel, this 4 and that 2 will cancel. And therefore, we’re going to end up with electric field is equal to *Q* over 2*πε*0*R* times 1 over 4*R*2 plus *L*2. Again, this electric field is in positive direction in vector form, therefore, we multiply that by unit vector, **î**.

Here, let’s go back to our example statement and let us use *R* instead of *a* here. Replace this *a* with big *R*, for the distance, so that we don’t have to replace this big *R* to *a*.

**So, the electric field of such a charged rod R distance away along its bisector is going to be equal to its quantity. So if we are dealing with a finite rod that we begin apply or follow exact the same procedure that we did for the infinite rod, the only difference is going to be that in our integral, we will substitute the appropriate boundaries.**