The first chapter of the sophomore physics Coulomb s law, the sophomore physics Coulomb s law questi

The premise of this problem is that when the balls of the same volume come into contact, the total amount of charge they carry will be divided equally.

The rest is easy to understand.

1) Assuming that the original charge is +1 and +1, C is charged + after the first contact A, and A is charged +, and B and C are both charged after the second contact (+1+, so A: B = +: +3

2) Assuming that the original charge is +1 and -1, C is charged + after the first contact A, and A is charged +, and B and C are both charged after the second contact (-1+, so A:B=+::1

Since the question asks about the amount of electric charge, positive and negative are not considered. So 2:3 and 2:1

This question is done according to the principle of first neutralization and then bisection after charge contact.

1) Let A and B both balls have the charge of Q, after C and A are in contact, C and A are both charged, then C and B are in contact again, altogether, after bisecting, the charge of each ball. Then the ratio of the charge of the two spheres A and B is: 3;

2) Let A with Q, B with Q, C and A are in contact, both have a charge, then C is in contact with B, the common charge, after bisecting, B and C are both charged, then the ratio of the charge of A and B is equal to: 1.

2) The contact charge of the two spheres is divided equally after c is in contact with a, the charge of a is +1 2, and after contact with b, the charge of b becomes -1 2, and the ratio of the charge is 1:1

I got it! According to the principle of charge distribution Ann "see one side and bury half with a smile"!

First, set the original power of A to +Q, and after C contacts A, the power of AC is Q 2 and Q 2 respectively!

The amount of electricity carried by CB after CB contact is:

q 2-q) 2 i.e. 1q tansheng grip 4

The electrostatic force between ab becomes 1 4 * 1 2 = 1 8 times!

If the distance is increased by a factor of 2, the Coulomb force is reduced by a factor of 4!

So let's celebrate 1 8*1 4=1 32

Charge c must be negatively charged and placed in the middle of ab, which can be understood by drawing a force analysis diagram.

Suppose the distance c from a is x and the amount of charge is qc

The repulsive force of b to a is fba=kq1q2 squared.

The attraction of C to A is FCA=KQ1QC x squaredTo make A balance, then FBA=FCA

Solution x 2q2=

In the same way, consider b

The repulsive force of A to B is Fab=kQ1Q2 squared.

The attraction of c to b is fcb=kq2qc (squared solution (

By De9(

The solution is x=, that is, c is 30cm away from a

Substituting into , we get qc=9q2 16

Point charge Coulomb force = q1*q2*k r 2

If it's the same kind of charge, it becomes a math problem, which is x+y=z, and find the maximum value of x*y.

This can be obtained by means inequality, when x=y, x*y is maximum, that is, when the two charges are equal, q1*q2, the maximum r k is unchanged, then the Coulomb force is the maximum, and it is much easier to get if it is different.

Both have less power, so the Coulomb force is definitely reduced.

If it is the same charge:

f1=kq^2/r^2

f2=k(q-δq)(q+δq)/r^2=k(q^2-δq^2)/r^2

f1 > f2 can be seen

In the case of a dissimilar charge, the absolute value of the charge becomes (q-δq), f2=k(q-δq) 2 r 2 after neutralizing a part of the δq charge

f1 > f2 can also be seen

So the interaction force between them is smaller compared to the original.

I'm sorry.,I got the question wrong.。。。 Upstairs is right!

1.This ball is placed between 2 small pies and the ball is lifted from +Q.

2.The third ball should be charged -9 16q

The analysis of the force on the ball makes the repulsion and gravity reach a balance.

Choose a small ball subject to gravity, tension, and Coulomb force.

The three forces form a vector triangular calendar ridge, this triangle is similar to the triangle formed by two ropes and two charges apart, the length of the two ropes is equal, so the pulling force of the rope is equal to the gravitational force.

When the limb ballast q is doubled and re-stabilized, the ball is still subjected to three forces, or it forms a vector triangle, or the triangle composed of two ropes is similar, the corresponding sides are proportional, and the length of the front and rear ropes remains the same, what changes is the charge spacing, and the pulling force is still equal to gravity.

Hello choose c because it is just satisfied.

Three charge balance mantras: "two same, one different, two big plus one small, near small and far-reaching".

That is to say, the third one is -, which is placed in the middle.

and q1q3= q1q2+ q2q3

If it is ab, then the left two are negative, no.

If d, then both on the right are positive, and it doesn't work.

Feel free to ask.

The answer is c, because each charge is balanced by the electrostatic force of the other two charges, so the charges on both sides have the same sign, and the different signs in the middle, and because the electricity of b is small, it must be on the outside of b.