How do i find the higher electric field of these two

[Dell]

given a ball and an infinite pipe, both with the same radius and same charge density $$\sigma$$
what is the total electric field at the meeting point of the 2 if we bring them close together so that they touch at point A?

what i did was:

the ball has spherical symetry and the pipe has cylindrical symetry, so those are the shapes i will use for my gauss law,

since R_A=R

for the ball
$$\Phi$$=EA=E(4$$\Pi$$R²)=$$\sigma$$(4$$\Pi$$R²)/$$\epsilon$$₀

E=$$\sigma$$/$$\epsilon$$₀

for the pipe
$$\Phi$$=EA=E(2$$\Pi$$R*L)=$$\sigma$$2$$\Pi$$R*L)/$$\epsilon$$₀

E=$$\sigma$$/$$\epsilon$$₀

does this mean that the total field at point A will be 0 since they have the same field in opposite directions?
does that make sense?

 

[Redbelly98]

Looks right to me.

 

[nlightner]

I would approach this problem by first understanding the concept of electric field and how it is affected by different factors such as charge density and distance. Then, I would use mathematical equations and principles such as Gauss's Law to determine the electric field at point A.

Based on the information provided, it seems that the ball and the pipe have the same radius and charge density, and they are brought close together so that they touch at point A. In this case, the electric field at point A will depend on the direction of the electric fields created by the ball and the pipe.

If the ball and the pipe have the same charge, but opposite signs, then the total electric field at point A will indeed be 0. This is because the electric fields created by the ball and the pipe will cancel each other out due to their opposite directions.

However, if the ball and the pipe have the same charge and the same sign, then the total electric field at point A will not be 0. In this case, the electric fields created by the ball and the pipe will add up, resulting in a higher electric field at point A.

In conclusion, to find the higher electric field at point A, we need to consider the charge and direction of the electric fields created by the ball and the pipe. This can be determined using mathematical equations and principles such as Gauss's Law.