1. The electrostatic force of attraction between two small spheres that are 1.0 meter apart is F. If the distance between the spheres is decreased to 0.5 meter, the electrostatic force will be
a) F/2
b) 2f
c) F/4
d) 4F
I know when the distance is decreased the Fe would decrease as well.
2. Two identical spheres, A and B, carry charges of +6 microcoulombs and -2 microcoulombs. If these spheres touch, what will be the resulting charge on sphere A?
1. What yu "know" is wrong. The attraction force increases when they get closer together. Have you heard of "inverse square law" behavior?
It applies here. Use it to figure out the answer.
2. The charges can move. The charges redistribute so that the total charge remains the same (4 uC) but each sphere has the same charge (because they are identical). That would have to be half of 4 uC.
k, thanks a lot. Have a good day :)
1. The electrostatic force is inversely proportional to the square of the distance between the spheres. So, when the distance is halved from 1.0 meter to 0.5 meter, the force will be (1/0.5)^2 = 4 times stronger. Therefore, the electrostatic force will be 4F. So, the answer is (d) 4F.
2. When the spheres touch, they will equalize their charges through the transfer of electrons. Since sphere B has a negative charge (-2 μC), it will transfer 2 μC of negative charge to neutralize sphere A. Thus, the resulting charge on sphere A will be +6 μC (original charge) - 2 μC (transferred charge) = +4 μC. Therefore, the resulting charge on sphere A will be +4 μC.
1. To answer the first question, we need to understand the relationship between the electrostatic force and the distance between two charged spheres.
According to Coulomb's Law, the electrostatic force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. The mathematical representation of Coulomb's Law is:
F = k * (Q1 * Q2) / r^2
where F is the electrostatic force, k is Coulomb's constant, Q1 and Q2 are the charges of the spheres, and r is the distance between them.
In this case, we have two small spheres that are initially 1.0 meter apart, and the electrostatic force between them is F. If we decrease the distance between the spheres to 0.5 meter, we can determine the new electrostatic force by using the inverse square relationship.
Using the inverse square relationship, if the distance is halved, the force will increase by a factor of 2^2 = 4. Therefore, the electrostatic force will be 4 times greater when the distance is decreased to 0.5 meter.
So, the correct answer is d) 4F, meaning the electrostatic force will be four times the initial force F.
2. To determine the resulting charge on sphere A when two identical spheres with charges of +6 microcoulombs and -2 microcoulombs touch, we need to understand the principles of charge conservation.
When two objects with opposite charges touch, electrons from the more negatively charged object will flow to the more positively charged object until both objects reach the same potential. This process is called charge neutralization.
In this case, sphere A has a charge of +6 microcoulombs, and sphere B has a charge of -2 microcoulombs. When they touch, some of the electrons from sphere B will transfer to sphere A until both spheres have the same charge. Since electrons have a negative charge, transferring electrons from sphere B to sphere A will reduce the overall charge of sphere A.
As a result, the charge on sphere A after the two spheres touch will be the difference between the initial charges:
(Charge on sphere A) = (Charge on sphere A initially) - (Charge on sphere B initially)
= +6 microcoulombs - (-2 microcoulombs)
= +6 microcoulombs + 2 microcoulombs
= +8 microcoulombs
Therefore, the resulting charge on sphere A will be +8 microcoulombs.
1. The electrostatic force of attraction between two small spheres that are 1.0 meter apart is F. If the distance between the spheres is decreased to 0.5 meter, the electrostatic force will be
a) F/2
b) 2f
c) F/4
d) 4F
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F = k/r^2
F = k/1^2
k = 1F for our purposes
Fnew = F/(.5)^2
Fnew = F/.25 = 4F