Two neutral metal spheres on wood stands are touching. A negatively charged rod is held directly above the top of the left spher
e, not quite touching it. While the rod is there, the right sphere is moved so that the spheres no longer touch. Then the rod is withdrawn. Afterward, what is the charge state of each sphere? Use charge diagrams to explain your answer.
A. Both the spheres are neutral.
B. The left sphere is negatively charged, the right sphere is charged positively.
C. The right sphere is negatively charged, the left sphere is charged positively.
D. Both the spheres are charged positively.
E. Both the spheres are charged negatively.
A. Both the spheres are neutral.
B. The left sphere is negatively charged, the right sphere is charged positively.
C. The right sphere is negatively charged, the left sphere is charged positively.
D. Both the spheres are charged positively.
E. Both the spheres are charged negatively.
1 answer:
Answer:
The right sphere is negatively charged, the left sphere is charged positively.
Explanation:
When a negatively charged rod is held above the top of left sphere, the rod will attract positive charges and repel negative charges. As the sphere are initially touching each other so positive charges from the both spheres will moves toward the rod. When we separate the spheres positive charges from right sphere have already moved toward the rod i.e. left sphere, creating a deficiency of positive charges in the right sphere and excessiveness of positive charges in left sphere , hence the right sphere will remain negatively charged and left sphere will remain positively charged.
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Answer:
Stay the same
Explanation:
First of all, let's find how the capacitance of the capacitor changes.
Initially, it is given by
where
is the vacuum permittivity
A is the area of the plates
d is the separation between the plates
From the formula, we see that the capacitance is inversely proportional to the separation between the plates. In this problem, the distance between the plates is doubled, so the capacitance will be halved:
The potential difference across the capacitor is given by
where
Q is the charge on the plates
C is the capacitance
We see that the voltage is inversely proportional to the capacitance. We said that the capacitance has halved: therefore, the potential difference across the two plates will double:
Now we can analyze the electric field between the plates of the capacitor, which is given by
we said that:
- The voltage has doubled: V' = 2 V
- The distance between the plates has doubled: d' = 2 d
therefore, the new electric field will be
So, the electric field is unchanged. And since the force on the particle at the center is directly proportional to the electric field:
F = qE
Then the force on the particle will stay the same.