Physics: Electric Fields

[kingwinner]

1) For a positively charged sphere, the electric field lines point outward. For a negatively charged sphere, the electric field lines point inward. Why is it always like that?

 

2) A charge (3.6x10^-6 C) is 30cm to the left of another charge (-2.7x10^-6 C). Point A is 20cm to the right of the charge with -2.7x10^-6.

a) Find the magnitude and direction of the net electric field at point A.

:lol: What electric force is exerted on a charge 4.5x10^-6 C placed at point A?

c) What electric force is exerted on a charge -4.5x10^-6 C placed at point A?

[For part a, I got an answer of 4.8x10^5 N/C

For part b, would the electric force have the same direction as the electric field at point A? (i.e. left?) I assumed so, and got an answer of -2.2N

The main problem is part c. I don't know how to calculate this. We know that E = Fe/q, where E represents the electric field, Fe is the electric force, and q is the magnitude of the charge at point A. Note that in the equation, q is substitute as a magnitude. So would the negative charge affect anything, will the answer be the same as part b?]

 

3) A negative charge of 2.4x10^-6 C experiences an electric force of magnitude 3.2 N, acting to the left.

a) Calculate the magnitude and direction of the electric field at that point.[The answer is 1.3x10^6N/C

. Why is the direction to the right? The electric force has a direction to the left, and since electric force and electric field always have the same direciton, shouldn't the direction of the electric field be to the left also?]

;) Calculate the value of the field at that point if a charge of 4.8x10^-6 C replaces the charge of 2.4x10^-6 C.

 

4) Consider the setup of two large, equally charge, parallel, flat conducting plates close together, the top plate positive and the bottom plate negative.

-The electric field is constant everywhere in the space between the parallel plates. [WHY?]

-The magnitude of the electric field at any point between the plates depends and is directly proportional ONLY on the magnitude of the charge on each plate. [Why is it so? I thought that E=Fe/q is the formula for electric field, shouldn't E be inversely proportional to q? (E=electric field, q=charge]

-The plate separation has no effect on the electric field. [WHY?]

 

Could anyone explain? I will really appreciate for your help! :wave:

[ronthepon]

For Q1, just remember that electric feild lines are paths that a unit positive charge will take when its put in the feild, provided that it cannot disturb the original feild significantly.

[sebbysteiny]

F = q1 q2 / r^2 * epsilon 0 if I remember right.

 

If q1 is -ve, then unless q2 is also -ve, F will be negitive. So if you replace one charge with an equal but opposite charge, you will get an equil but opposite force.

 

So yes, ANS c = - ANS b

 

3) again, if a -ve charge will go to the left, a +ve charge will go to the right so the E field also goes to the right.

 

4) Both plates will be at the same VOLTAGE. Thus, for the potential energy at every part of the plate to be even, the charge density must also be even. Thus, by symetry, the E field is constant between the plates (ignoring edge effects).

 

Will think about the others.