point charge question?

Question

Two identical conducting spheres carry charges of +5.4 µC and -1.4 µC respectively. They are initially a large distance L apart. The spheres are brought together, touched together, and then returned to their original separation L. What is the ratio of the magnitude of the force on either sphere after they are touched to that before they were touched?

Answer 1

First, let's see what happens when we touched the spheres together. The first sphere carries more net charge than the other one. The negative charge on the second sphere cancel with a part of the positive charge on the first sphere, so both spheres are left with +5.4 - 1.4 uC = 4.0 micro coulombs. Since the spheres are identical (meaning they have the same size and therefore the same surfice area) the left charge is equally distributed among both spheres - each sphere gets half of the left charge, so +2.0 micro coulombs.
Then we have to remember that, according to Coulomb's law, electric force between two point charges is directly proportional to the product of the charges (it also depends on the separation, but this remains the same, so it's not important). The magnitude of the force is the same for both spheres (3rd Newton's law). We're looking for the ratio, so we just have to find the ratio of the charge products after and before the event:
2.0 * 2.0 / 5.4 * 1.4 = 0.529

Answer 2

Before they touched the force was proportional to the product of the charges (and it's attractive)

After they touch, the net charge (4 microcouloms), spreads evenly, so each gets +2 microcoulombs.

After, the force is proportional to the product of the two charges (and it's repulsive).

So the ratio is -(5.4)(1.4)/2/2

Answer 3

The electrostatic force is proportional to q1*q2/(r*r). The before and after, r is the same, so the ratio of force before and after is simply the ratio of q1 (before)*q2 (before) to q1(after)*q2(after).

The question is what is q1(after) and q2(after). Because they touch, each sphere split the net charge afterwards, ie 1/2 (5.4+ -1.4) = 1/2 (4) =+2 uC each.

So F (before)/F(after) = 5.4*(-1.4)/(2*2)

This gives the ratio of force as well as direction. F(before) is attractive while F(after) is repelling

Answer 4

After touching the charges balance so that charge on each = (5.4-1.4)/2 = 2.0

Force before proportional to q1q2 = 5.4 x 1.4
Force after proportional to 2.0 x 2.0

Work out 2.0 x 2.0 / (5.4 x 1.5)

Answer 5

volume charge density is the charge according to unit volume of a volume charge distribution. a factor charge is a discrete charge won't have a volume charge density (basically think of in case you candetermine the size section or volume of a 'factor'). And the formulation E = (one million / 4??) (Q/r²)r^, (the place r^ is the unit vector interior the process the unit beneficial try charge wrt to the factor charge Q) is the definition of the electrical powered field intensity. you have not got a evidence for that.