Electric Field Due To A Hemisphere

It is obvious that the direction vector will lie in the $xz$ plane, charge distribution is symmetric about the y-axis, and hence the $\hat j$ component will cancel out.

Considering the direction vector to be $a\hat i+b\hat k$ , we can say that the direction of the field of a similar hemisphere placed such that it is a mirror image of the current hemisphere along the z-axis, will be $a\hat i-b\hat k$ , and it will have the same magnitude.

Now looking at the two hemispheres as one complete spherical shell, we can say that the net field at that point is $0$ , i.e. the sum of fields due to the two hemispheres should be $0$ . Thus $Ea\hat i+Eb\hat k + Ea\hat i - Eb\hat k=0$ , Where $E$ is the magnitude of the field.

Hence $2Ea\hat i=0$ $a=0$

So it is obvious that the direction should be $-\hat k$ .

Well, you could easily do this by using symmetry.

Consider a sphere, and divide it into two hemispheres where one is defined as in the question.

Let's say that, $E_{1}$ is the field due to the above defined hemisphere, and $E_{2}$ is the field due to the complement hemisphere.

$E_{1}= (E_{1x},E_{1y},E_{1z})$

and, $E{2}= (E_{2x},E_{2y},E_{2z})$

In general, on any point on the x axis, $E_{y}=0$ (By symmetry).

and, $E_{1z} = -E_{2z}, E_{1x}=E_{2x}$

So if we superimpose $E_{1}$ and $E_{2}$ then it becomes a field inside a spherical shell, which is zero. Thus,

$E_{1x}+E_{2x}=0 \Rightarrow 2*E_{1x}=0 \Rightarrow E_{1x}=0.$

So the direction is -k.

What I did was think of how if it was a full shell, there would be no electrical field inside the sphere (as everything would cancel out). That helped me understand that the net electrical field can only be in one direction because when the other half/hemisphere is connected, only one directional sum/vector sum will change (because all the others will be canceled out).

So if you just picture it in 3D, if you were to measure the electrical field at $-2\hat{i}$ , the electrical field along the $\hat{z}$ and $\hat{i}$ directions would end up being canceled out, and the only thing left would be an electrical field pointing along the $-\hat{k}$ direction.

By the way, this is all possible because of the uniform charge distribution and the symmetrical shape.

Use principle of superposition. Assume the hemisphere is positively charged. Further consider a "reflection" or a carbon copy of the hemisphere brought under the original hemisphere. That will give us a full spherical shell. And remember that electric field inside spherical shell (conducting or not) is 0!

Thus by principle of superposition the sum of electric fields caused by the original hemisphere and it's reflection must be 0. Also note that the electric field caused by the "carbon copy" hemisphere must be a vector which is the reflection of the original electric field vector. Bottom line:-

1-The two Electric Field vectors are reflection of each other. 2-their Vector sum is null vector.

Thus they have to be along a line which is perpendicular to the line joining the centre to the given point. Which gives the answer -k (for positively charged hemisphere).