Electric Force problem -> Infinite charged plane with hole

[moonrkr]

Electric Force problem --> Infinite charged plane with hole

The plane is infinite charged. It has a charge density (σ) of 10nC/m$^{2}$. If R=5cm, determine the electric force of a proton in the point P=(0,0,10cm).

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MY set UP:
I was thinking about using E=σ/2*ε$_{o}$
and use F=Eq.
I can see problems in the book with the infinite charged plane, but they don't have a hole thru it... PLEASE HELP>>>!

 

[mathfeel]

The plane is infinite charged. It has a charge density (σ) of 10nC/m$^{2}$. If R=5cm, determine the electric force of a proton in the point P=(0,0,10cm).

=====================================================================



MY set UP:
I was thinking about using E=σ/2*ε$_{o}$
and use F=Eq.
I can see problems in the book with the infinite charged plane, but they don't have a hole thru it... PLEASE HELP>>>!

Consider the hole as a combination of positive and negative charges.

 

[DiracRules]

You can face this problem at least in two ways:
1) Calculate explicitly the force with the Coulomb expression $\vec{F}=\frac{q_1q_2}{4\pi\epsilon r^2}$
2) Solve the problem "geometrically": how do you build a charged plane with a hole? You can think either $\vec{F}_{positively\,charged\,plane\,with\,hole}=\vec{F}_{positively\,charged\,plane\,without\,hole}-\vec{F}_{field\,of\,the\,hole}$ or $\vec{F}_{positively\,charged\,plane\,with\,hole}=\vec{F}_{positively\,charged\,plane\,without\,hole}+\vec{F}_{negatively\,charged\,hole}$

Both ways, you should be careful, because the symmetry of the problem allows you to do powerful simplifications on the components of the forces acting on the proton.

 

[moonrkr]

Force of the hole - Force of the plane (without hole) = Force of the entire setting(plane with hole)?

 

[paranormal]

Based on the given information, we can calculate the electric field at point P using the formula E=σ/2*ε_{o}. This will give us the electric field at point P due to the infinite charged plane. However, since there is a hole in the plane, the electric field at point P will be zero due to symmetry. This means that the electric force on a proton at point P will also be zero.

In general, the electric force on a charged particle in an electric field is given by F=Eq, where E is the electric field and q is the charge of the particle. Since the electric field at point P is zero, the electric force on the proton will also be zero. This is because the electric field is a vector quantity and its direction determines the direction of the electric force. In this case, the electric field is perpendicular to the direction of the proton, so the electric force will be zero.

It is important to note that the electric field and electric force at point P will be non-zero if the hole in the plane is not symmetrically placed. In that case, the electric field will not cancel out and the electric force on a charged particle at point P will be non-zero. However, in the given scenario, the electric force on a proton at point P is zero due to the symmetry of the plane.