What is the direction of the E field at point P due to the sphere?

In summary, the problem involves an infinite plane charged with σ = 2μC/m^2 parallel to the yz plane at x = 2m, and a sphere with a radius of 0.8m and a charge density of ρ = -6μC/m^3 centered at x = 1m, y = 0m. The task is to find the intensity of the electric field at x = 1m and y = 0.5m. By using Gauss's law, the electric field at P can be found by adding the electric fields of the plane and the sphere, with the direction of the sphere's electric field at point P being determined by symmetry. To find the unit vector r
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Homework Statement


an infinite plan,parallel to the yz plan which passes from x=2 meter, is charged equally with the σ= 2μC/m^2 .A sphere with a radius 0.8 meters has its center in the in x=1 m,y=0 and has ρ=-6μC/m^3.Find the value of the intensity of the electric field in the point x=1 m and y=0.5 m.



Homework Equations


the figure :
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The Attempt at a Solution


The E at P is = Eplane + Esphere
I find the E of plane =(-σ/2∈ )* i=-113kN/C * i
To find the E of sphere, I have to use the formula E=(4*pi/3)*k*r*ρ*rˆ... I have every known data,I just need to find the unit vector r or (rˆ)..how to find it?
 
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  • #2
By symmetry, what is the direction of the E field at point p due to the sphere by itself?
Hint: invoke Gauss's law.
 

FAQ: What is the direction of the E field at point P due to the sphere?

1. What is a Gauss surface?

A Gauss surface is a hypothetical surface used in physics to calculate the electric field and electric flux of a charged object. It is a closed surface that can be any shape, and is often chosen to simplify calculations.

2. How is a Gauss surface related to Gauss's Law?

Gauss's Law states that the electric flux through a closed surface is directly proportional to the enclosed charge. A Gauss surface is used to apply this law and determine the electric field at a point due to a charged object.

3. Can a Gauss surface be open?

No, a Gauss surface must be closed in order to apply Gauss's Law and calculate the electric field and flux. An open surface would result in an infinite and unrealistic electric field.

4. What is the purpose of using a Gauss surface?

A Gauss surface is used to simplify calculations of the electric field and flux of a charged object. It allows us to apply Gauss's Law, which can be a more efficient method than directly calculating the electric field at every point.

5. How do you choose a Gauss surface for a specific problem?

The choice of a Gauss surface depends on the symmetry of the problem. The surface should be chosen so that the electric field is constant and perpendicular to the surface at every point. This allows for easier integration and calculation of the electric field and flux.

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