An uncharged spherical conducting shell surrounds a charge -q at the center of the shell. The charges on the inner and outer surfaces of the shell are respectively...
a] -q,-q
b]+q, +q
c]-q, +q
d]+q,0
e]+q, -q
2) A postive point Charge is enclosed in a hollow metallic sphere which is not grounded. At a point directly above the hollow sphere the elcetric field caused by the enclosed positive charge...
a]diminished to zero
b]diminished somewhat
c]increased somewhat
d] no change
3]A point charge +Q is located on the x-axis at x=a and a second charge -Q on x-axis at x=-a. A Gaussian surface with radius r=2a is centered at the origin. The flux through that surface....
a]0 b/c the negative flux over one hemishpere is equal to the positive flux over the other.
b]< 0
c] 1 b/c at every point on the surface the electric field has no comp. perpendicular surface
d]1 bc the electric field is 1 at every point on the surface.
e]None
mjvhyubhu1
1) The charges on the inner and outer surfaces of the shell are determined by the principle of electrostatic equilibrium. In this case, the charge -q at the center of the shell induces a charge distribution on the inner and outer surfaces of the shell that ensures the electric field within the shell is zero.
Since the charge inside the shell is -q and the electric field within the shell is zero, the induced charge on the inner surface of the shell must be +q to cancel out the electric field due to -q. Therefore, the charges on the inner and outer surfaces of the shell are +q and -q, respectively.
So, the correct answer is: c) -q, +q.
2) The electric field caused by a charge enclosed within a perfectly conducting hollow sphere is zero at all points within the hollow sphere. This is a consequence of the principle of electrostatic equilibrium in conductors.
Since the electric field within the hollow sphere is zero, it will remain zero at any point outside the hollow sphere. Therefore, the electric field caused by the enclosed positive charge will not change at a point directly above the hollow sphere.
So, the correct answer is: d) no change.
3) The flux through a Gaussian surface is given by the formula Φ = q_enclosed / ε₀, where q_enclosed is the charge enclosed by the Gaussian surface and ε₀ is the permittivity of free space.
In this case, the Gaussian surface has a radius of 2a and is centered at the origin. The charges +Q and -Q are both located on the x-axis. The charge enclosed by the Gaussian surface is the sum of the magnitudes of +Q and -Q, which is 2Q.
Since there is no charge within the Gaussian surface, the electric field is zero at every point on the surface. Therefore, the flux through the surface is zero.
So, the correct answer is: a) 0.
1) In this scenario, the uncharged spherical conducting shell surrounds a charge -q at the center of the shell. Since the shell is conducting, the charges on its inner surface will arrange themselves in such a way that they cancel out the electric field inside the shell. Thus, the inner surface of the shell will have a charge of +q.
On the other hand, the outer surface of the shell will not be influenced by the charge at the center, since the electric field inside a conducting shell is zero. As a result, the charge on the outer surface will also be +q.
So, the answer is option b) +q, +q.
To determine this, you can apply the principle of charge distribution on a conducting shell, which states that the electric field inside a conductor is zero, and any excess charge resides on the outer surface of the conductor.
2) In this situation, a positive point charge is enclosed in a hollow metallic sphere, which is not grounded. The electric field caused by the enclosed positive charge will remain the same at any point outside the sphere.
This is because the electric field due to a point charge depends on the magnitude of charge and the distance from the charge, but it is not affected by the presence of a conducting sphere (which does not dissipate charge or change the electric field). Therefore, the answer is option d) no change.
To determine this, you can use the principle that the electric field outside a charged conducting sphere does not depend on the charge distribution inside the sphere, but only on the total charge enclosed by the sphere.
3) In this scenario, there are two point charges: +Q at x = a and -Q at x = -a. A Gaussian surface with radius r = 2a is centered at the origin.
The flux through this Gaussian surface can be determined using Gauss's Law, which states that the flux of the electric field through a closed surface is proportional to the net charge enclosed by the surface divided by the permittivity of free space.
In this case, the Gaussian surface encloses both the +Q and -Q charges, resulting in a net charge of 0. Since the net charge enclosed is 0, the flux through the surface will also be 0.
So, the answer is option a) 0.
To arrive at this conclusion, you can apply Gauss's Law and consider the fact that the electric field due to +Q and -Q charges at the Gaussian surface will have equal magnitudes and opposite directions, canceling each other out.
Therefore, the correct answer for each question is:
1) b) +q, +q
2) d) no change
3) a) 0