Electric Charges And Fields

Step 1: Field inside the sphere. Inside a uniformly charged sphere, electric field varies directly with distance. E r
Step 2: Field outside the sphere. Outside the sphere, electric field behaves like that of a point charge. E (1)/(r²)
Step 3: Graph interpretation. The field increases linearly till R and then decreases as 1/r².

Step 1: Balance of forces.
The force due to gravity on an electron is , and the force due to electric field is . To balance the gravitational force, we set: where is the mass of the electron, is the acceleration due to gravity, and is the charge of the electron.

Step 2: Solve for electric field intensity.
Substitute the values of , , and to get the electric field intensity . Thus, the correct answer is (1).

Step 1: Electric field of a dipole on its axial line:
Step 2: Calculate fields due to both dipoles at the given point and add vectorially.
Step 3: Resultant field is along direction with magnitude:

Step 1: Charge density varies as .
Step 2: Charge enclosed within radius :
Step 3: Electric field inside sphere:
Step 4: Given:
Step 5: Comparing powers: (Considering volume charge proportionality, corrected value gives ).

or

For
Given
or

or .

Step 1: Use Gauss’s law for different regions.
Step 2: - Inside shell A: E=0 - Between A and B: field due to +2q - Outside both shells: net charge = +q

Step 1: The electric potential due to a point charge q at a distance r is: V = (kq)/(r)
Step 2: All four charges are placed at the corners of a square, so their distances from the centre are equal. Hence, the net potential at the centre is proportional to the algebraic sum of the charges.
Step 3: Sum of charges: (-Q) + (-q) + 2q + 2Q = Q + q
Step 4: For the potential at the centre to be zero: Q + q = 0 ⟹ Q = -q

Step 1: Dipole moment: vecp = int vecrdq
Step 2: By symmetry, vertical components cancel.
Step 3: Net dipole moment lies along x-axis: p = (2Rq)/(π)

Step 1: Charge enclosed within radius r: Qᵣ ∝ int₀ʳ krᵃ r² dr ∝ rᵃ+3
Step 2: Electric field inside the sphere: E(r) ∝ (Qᵣ)/(r²) ∝ rᵃ+1
Step 3: Given condition: (E(R/2))/(E(R)) = ((1)/(2))ᵃ+1 = (1)/(8)
Step 4: Solving, a+1 = 3 ⟹ a = 5

Step 1: Electric field on the axis of a uniformly charged disc at distance x: E = (σ)/(2varepsilon₀) (1 - fracx√(x² + R²))
Step 2: At the centre (x=0): E₀ = (σ)/(2varepsilon₀)
Step 3: At x = R: ER = (σ)/(2varepsilon₀) (1 - \frac1√(2))
Step 4: Fractional reduction: (E₀ - ER)/(E₀) = \frac1√(2) ≈ 0.293
Step 5: Percentage reduction: 29.3%

If a dipole is placed inside a sphere, the electric flux through the sphere will depend on the orientation of the dipole with respect to the surface of the sphere.
The electric flux through a closed surface is given by Gauss's law, which states that the total electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of the medium.
If the dipole moment is aligned with the normal vector of the surface, the electric flux through the sphere will be zero. This is because the positive and negative charges of the dipole cancel each other's contributions to the electric field at every point on the surface, resulting in no net flux passing through. If the dipole moment is not aligned with the normal vector of the surface, the electric flux through the sphere will be nonzero. In this case, the positive and negative charges of the dipole do not cancel each other completely, resulting in a net electric field passing through the surface of the sphere. This net electric field will lead to a nonzero electric flux through the sphere.
Therefore, the electric flux through the sphere with a dipole inside will depend on the orientation of the dipole relative to the surface of the sphere.