Electric Field and charged plane

In summary, an infinite, charged plane/plate with a uniform positive charge density of σ exerts an electric field of E=σ/2ε on point P, which is positioned between the plane and a distant positively charged particle q. To find the distance between point P and particle q, the electric field exerted by the plane and particle q is set equal to 0, and the distance r is calculated. In part B, the plane is replaced with a new positively charged particle Q, and the value of Q at point A is calculated using the equation E=k * Q/r^2, where r is the distance found in part A.
  • #1
Hitchslaps
6
0

Homework Statement



An infinite, charged plane/plate has a uniform positive charge density of σ. Another positively charged particle is found at a distant of D from the plane. In point P, positioned between the two, the electric field equals 0.
A. What is the distance between point P and the charged particle q?
B. The plane is removed and replaced with a new positively charged particle Q. What should be the value of Q in point A in order for the electric field to remain 0?

Homework Equations



E=σ/2ε , E=k * Q/r^2

The Attempt at a Solution



In A, I've calculated the electric field the plane exerts on point p, which is E=σ/2ε, and then added the electric field exerted by particle q, which is E=k * Q/(D-r)^2.
E=k * Q/(D-r)^2 + σ/2ε =0
then found r. Am I right on this? Please help.

In B, just use this equation E=k * Q/r^2 instead of E=σ/2ε, correct?
 

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  • #2
Hi Hitch, and welcome to PF.

Good use of the template, clear statements and I think your attempt is good. I wouldn't approach it any different -- although: Did you find an imaginary r ? Because everything I see in the expression looks positive...

Tackling B in the way you propose is fine, too.
 
  • #3
Thanks :)

How would you find the distance from point P to the particle q? I've thought of another way: maybe use a new variable such as x, and then subtract x from D, as in E=k * Q/x^2 ; distance from p= D-x?
I'm unsure.
 
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  • #4
You have already found this distance , namely in part A. You called it r.
Now you want to calculate what charge is needed to create the same field at point P as the plate did. Not so difficult!
 
  • #5
So in B I use E=k * Q/(D-r)^2 with r, the distance I found in A. Thanks again!
 

1. What is an electric field?

An electric field is a physical quantity that describes the effect of electric forces on charged particles. It is defined as the force per unit charge acting on a charged particle at any given point in space.

2. How is an electric field created by a charged plane?

An electric field is created by a charged plane when the plane has a non-uniform distribution of charge. The electric field lines are perpendicular to the plane and point away from positive charges and towards negative charges.

3. What is the difference between an electric field and an electric potential?

An electric field is a vector quantity that describes the direction and strength of the force on a charged particle, while electric potential is a scalar quantity that describes the amount of work needed to move a charged particle from one point to another in an electric field.

4. How do you calculate the electric field strength of a charged plane?

The electric field strength of a charged plane can be calculated using the equation E = σ/2ε₀, where σ is the surface charge density and ε₀ is the permittivity of free space.

5. What is the significance of a charged plane in practical applications?

A charged plane is commonly used in practical applications such as capacitors and parallel plate electrodes in electronic devices. It is also used in electrophoresis, a technique used in biochemistry and molecular biology to separate and analyze DNA, RNA, and proteins.

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