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# E-field problem

1.

The surface area of a sphere is A=4pi r^2,

and the charge density on the sphere surface is q/A.

Hence,

E=charge density on the sphere surface/e.(where e denotes permittivity)

This formula is true in sphere,but is it also true in two parallel plates?

Why?Please explain it.Is this formula hold for any shape?

2.

Why don't we directly use E=q/4pi e r^2 instead of Gauss's law to prove that E-field inside a conductor is zero?I think by E=q/4pi e r^2,inside the sphere,q=0,then E=0.

What wrong with my concept?

3.

In the case of two parallel plates,

The book claim that,

E=-dv/dx

E=potential difference/separation between two plates

E=V/d

In the second line,why is the minus sign disappear?

One more question:

How to prove the E-field between two parallel plates is constant instead of just looking at the field line?

1.So,what area should be took?Which E-field should be took?both the 2 plates of what?Could you show the proof?

2.That nice.OK

3.In E=V/d,is that V(potential different) means the work done for a positive charge to bring from negative plate to positive plate?

4.That nice.OK

### 3 Answers

- 天同Lv 71 decade agoFavorite Answer
1. Certainly, the formula you mentioned is not generally true. It is valid for a sphere and a parallel plate arrangement because the charge distribution on these devices is uniform. For non-uniform distribution of charges over the surface of an object, the electric field intensity on the surface varies with location. You need to do the integration of the field intensity with different unit areas over the entire object surface.

2. The equation E = q/4.pi.e.r^2 gives the value of electric field intensity E outside a charged spherical conductor, i.e. for r > radius of sphere. In fact, the contradiction of results (q =/= 0 coul on sphere, but E = 0 v/m inside sphere) illustrates that the equation cannot be applied.

3. The negative sign (not minus sign) indicates that the potential decreases along the direction of the field lines.

In the parallel plate arrangement, if we go along the field lines from the positive plate, say of potential Vp, to the negative plate, say of potential Vn, the electric field intensity E is given by,

E = - (Vn - Vp)/d, where d is the plate separation

i.e. E = (Vp - Vn)/d

since Vp > Vn, the quantity (Vp - Vn)/d is positive.

Q: How to prove the E-field between two parallel plates is constant instead of just looking at the field line?

It is simply that the charge distribution on any one of the plates is uniform in view of the plates are flat pieces of conductor. You cannot differentiate physically an unit are on one part of the plate with those on other parts.

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- ChauLv 51 decade ago
Would you mind to show the step?

2010-02-04 00:03:20 補充：

OK.Thank you!Actually,I am a F.6 student.

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- Prof. PhysicsLv 71 decade ago
For Q.1, you have to use Gauss' Law

2010-02-04 00:01:13 補充：

Are you a high school student or a university student?

You may consult some university Physics textbook. Anyone will do.

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