Electrostatics

Step 1: Understanding the problem.
We have a charged sphere with charge density , and a smaller sphere is cut out from it. The question asks for the electric field at a point due to the charge distribution, considering the symmetry of the setup.
Step 2: Applying Gauss's Law.
We can use Gauss’s law to solve for the electric field at a point outside the charged sphere. First, we know that the electric field due to a uniformly charged sphere is given by: where is the enclosed charge, and is the distance from the center. After the smaller sphere is cut out, we can treat the remaining charge as a net charge distribution.
Step 3: Considering the cut-out sphere.
The electric field at is due to the charge left in the sphere after cutting out the smaller sphere. The magnitude of the electric field is related to the net charge left after the cut-out.
Step 4: Conclusion.
By solving the equations, the value of is determined to be 6. Thus, the electric field at is .

Step 1: Understanding the electric potential in a coaxial cable.
The electric potential in a coaxial cable is determined by the radial distance from the center of the cable. The general form of the potential in a coaxial cable is given by: where is the potential difference between the inner and outer conductors, is the radius of the inner conductor, and is the radius of the outer conductor.
Step 2: Substituting the known values.
Substitute the known values:
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The potential at is calculated as: which gives a result of 8 V.
Step 3: Conclusion.
Thus, the potential at a distance of 3.5 mm from the axis is 8 V.

Step 1: Understanding the dipole electric field.
The electric field of a dipole varies with distance and direction. The work done in moving a charge in the field of a dipole is determined by the interaction between the dipole's electric field and the charge. The work done in moving a charge is given by , and the potential at is zero for the dipole.
Step 2: Conclusion.
Thus, the correct answers are options (B) and (C).

Step 1: Understanding the problem.
The problem involves a sphere with uniform charge density , from which a smaller sphere is cut out. The electric field at a point inside the sphere, after the cut, needs to be calculated. Step 2: Electric field due to a uniformly charged sphere.
The electric field inside a uniformly charged sphere with charge density is given by: where is the radial distance from the center of the sphere. Step 3: Applying the cut-out sphere concept.
Since a smaller sphere of radius is cut out from the original sphere, the electric field at the point where the cut was made must account for the effect of this absence. This results in the electric field expression: Therefore, the integer in the equation is 8.

Step 1: Use the potential formula for coaxial cables.
The potential at a radial distance from the axis in a coaxial cable is given by: where and are the radii of the inner and outer conductors respectively, and is the point where the potential is to be calculated. Step 2: Apply the given values.
Here, , , and , with the inner conductor at 10 V and the outer conductor grounded. Step 3: Calculation of potential.
By applying the formula and solving for the potential at , the calculated value of the potential is between 3.79 and 3.99 V.

Step 1: Understanding the dipole electric field.
The electric field of a dipole varies with distance and direction. The force on a charge placed near a dipole is determined by the dipole field and the interaction between the dipole moment and the charge. The force on the charge is given by .
Step 2: Conclusion.
Thus, the correct answer is option (D).

Step 1: Compute enclosed charge for .
. Total charge inside radius is

Step 2: Compute total charge of the sphere.
Thus

Step 3: Substitute into .

Step 4: Apply Gauss's law.
.
Thus

Step 5: Conclusion.
Electric field for is

Step 1: Use Gauss's law in differential form.
Compute divergence: Both terms are nonzero because derivatives of exponentials exist. Thus .

Step 2: Interpretation.
If divergence is nonzero, charge density is nonzero. Thus region contains a non-uniform charge distribution.

Step 3: Conclusion.
Correct: (A).

Step 1: Recall field of an infinite sheet.
An infinite plane with charge density produces electric field

directed away from the positive sheet and toward the negative sheet.

Step 2: Locate the point relative to the sheets.
Point is at .
Sheets are at (negative), (positive ), (positive ).

Step 3: Determine direction of each field.
- For sheet at (negative charge): field at points toward the sheet ⇒ downward (− ). Magnitude .
- For sheet at (positive ): point is below the sheet ⇒ field downward (− ). Magnitude .
- For sheet at (positive ): point is below the sheet ⇒ field downward (− ). Magnitude .

Step 4: Add magnitudes (all downward).
Total field:
downward.
downward = (since downward is + in their diagram).

Step 5: Conclusion.
The electric field at is .

Step 1: Formula for electrostatic energy of a charged spherical shell.
The electrostatic energy of a uniformly charged spherical shell is given by: where is the surface charge density, and is the radius of the shell. Since the total charge is 5 C, the surface charge density is:
Step 2: Calculate the electrostatic potential energy.
Using the formula for the energy stored in a spherical shell: where is the radius of the shell. Substituting the values:
Step 3: Convert the answer to the desired units.
Finally, in terms of the given units: Thus, the correct answer is 1.781 .

Step 1: Formula for electric displacement vector .
The electric displacement vector due to a line charge is given by: where is the charge density, , and is the permittivity of free space.
Step 2: Calculate the electric displacement vector.
Substitute the known values:
Step 3: Conclusion.
The magnitude of the electric displacement vector at a point 20 mm away from the axis of the wire is approximately 1.131 C/m².

Step 1: Understanding properties of perfect conductors.
A perfect conductor is a material in which the free electrons move without resistance. In electrostatic equilibrium, several key properties hold: 1. The surface of the conductor is an equipotential surface because the electric field inside the conductor is zero. 2. Any net charge on a conductor resides on the surface, as charges move to the surface to cancel out the electric field inside. 3. The electric field is zero inside the conductor because the free electrons rearrange to cancel any internal electric field. 4. Just outside the conductor, the electric field is perpendicular to the surface, as the electric field lines cannot have a component parallel to the surface in electrostatic equilibrium.

Step 2: Analyzing the options.
(A) The conductor has an equipotential surface: Correct. The entire surface of a conductor is an equipotential in electrostatic equilibrium.
(B) Net charge, if any, resides only on the surface of conductor: Correct. Any excess charge resides on the surface of the conductor.
(C) Electric field cannot exist inside the conductor: Correct. The electric field inside a conductor is zero in electrostatic equilibrium.
(D) Just outside the conductor, the electric field is always perpendicular to its surface: Correct. The electric field just outside a conductor is normal (perpendicular) to the surface.

Step 3: Conclusion.
All the statements are correct, so the correct answer is (A), (B), (C), (D).

Step 1: Understanding the forces involved.
The spherical ball with charge experiences two main forces: - The gravitational force , acting downward. - The electrostatic force due to the surface charge on the sheet, which creates an electric field . The electric field due to a uniformly charged infinite sheet is given by: where is the surface charge density of the sheet and is the permittivity of free space.
Step 2: Forces on the spherical ball.
At equilibrium, the forces on the spherical ball are balanced. The ball experiences the following: - The electrostatic force , acting to the right (towards the sheet). - The gravitational force , acting downward. - The tension in the string, which has two components: one vertical (balancing ) and one horizontal (balancing ). Since the string makes a 45° angle with the sheet, we can resolve the tension into horizontal and vertical components:
Step 3: Solving for .
Using the fact that , we can equate the horizontal and vertical components of tension: From the vertical component: From the horizontal component: Simplifying: Thus, the correct expression for is:
Step 4: Conclusion.
The correct answer is (D) .

Step 1: Bound charge densities.
The surface bound charge density is given by At , the normal is inward, so . At , the normal is outward, so .

Step 2: Volume bound charge density.
The volume bound charge density is Since , Thus, .

Step 3: Conclusion.
Hence, surface charge densities are and , and there is no volume bound charge inside.

Step 1: Understanding the concept.
The electrostatic potential energy of a system of charges is the sum of the potential energies for all distinct pairs of charges. For two point charges and separated by a distance , the potential energy is

Step 2: Apply to the given system.
In an equilateral triangle, all three sides are equal to . There are 3 distinct pairs of charges: (1,2), (2,3), and (3,1). Hence, the total potential energy is the sum of potential energies of these three pairs:

Step 3: Simplify the expression.

Step 4: Conclusion.
The electrostatic potential energy of three identical charges at the vertices of an equilateral triangle is .