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2 answers:
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Answer:
A
Explanation:
The figure shows the electric field produced by a spherical charge distribution - this is a radial field, whose strength decreases as the inverse of the square of the distance from the centre of the charge:
More precisely, the strength of the field at a distance r from the centre of the sphere is
where k is the Coulomb's constant and Q is the charge on the sphere.
From the equation, we see that the field strength decreases as we move away from the sphere: therefore, the strength is maximum for the point closest to the sphere, which is point A.
This can also be seen from the density of field lines: in fact, the closer the field lines, the stronger the field. Point A is the point where the lines have highest density, therefore it is also the point where the field is strongest.
1-2) They have same surface charge density
3-4) The metallic conductor has greatest surface charge density
Explanation:
1-2)
In a conductor, the charge carriers (mainly electrons) are free to move. Therefore, as a result, they tend to move at the largest possible distance from each other, because of the repulsive force that they exert on each other.
The configuration that maximize the distance between the charge carriers for a solid sphere of metallic conductor is the one in which all the electrons are on the surface, and they are equally spaced between each other. This means that for the solid sphere of radius R, the excess charge Q will be entirely spread over the surface of the sphere.
Similarly, the excess charge Q on the hollow spherical shell (which is also made of the same conducting material) will also be spread over the surface with the charge carriers at the maximum distance from each other. Therefore, the surface charge density for both objects will be
where R is the radius of the two spheres.
3-4)
In this case, the surface charge density on the two objects is different.
In fact, on the metallic sphere (conducting) the surface charge density is (as explained in part 1):
Hoever, the second sphere is made of an insulating material. In an insulator, the charge carriers are not free to move. If the initial charge Q is spread across the all sphere (which is not hollow), this means that some of the charge will actually also be inside the sphere. So the charge deposited on the surface, Q', will be less than the total charge Q. Therefore, the surface charge density will be
And since Q' < Q, this means that , so the conducting sphere has a greatest surface charge density.
Complete Question
An isolated charged soap bubble of radius R0 = 7.45 cm is at a potential of V0=307.0 volts. V0=307.0 volts. If the bubble shrinks to a radius that is 19.0%19.0% of the initial radius, by how much does its electrostatic potential energy ????U change? Assume that the charge on the bubble is spread evenly over the surface, and that the total charge on the bubble r
Answer:
The difference is
Explanation:
From the question we are told that
The radius of the soap bubble is
The potential of the soap bubble is
The new radius of the soap bubble is
The initial electric potential is mathematically represented as
The final electric potential is mathematically represented as
The initial potential is mathematically represented as
The final potential is mathematically represented as
Now
substituting values
=>
So
Therefore
where k is the coulomb's constant with value
substituting values