E-field problem

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.

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.

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.