Answer :
a. i. Gauss's theorem in electrostatics, also known as Gauss's law, states that the total electric flux (Φ_E) through any closed surface in a vacuum is equal to the net electric charge (Q_enc) enclosed within that surface divided by the electric constant (ε_0), which is the permittivity of free space. Mathematically, it is expressed as:
Φ_E = Q_enc / ε_0
ii. Electric flux is the measure of the flow of the electric field through a given area. It is defined as the product of the electric field (E) and the area (A) perpendicular to the field, and it is a scalar quantity. Mathematically, electric flux (Φ_E) through an area is given by:
Φ_E = E · A · cos(θ)
where θ is the angle between the electric field lines and the normal (perpendicular) to the surface.
b. To calculate the electric field intensity (E) due to a line charge using Gauss's theorem, we follow these steps:
1. Choose an appropriate Gaussian surface. For a line charge, a cylindrical surface co-axial with the line charge is appropriate because of the symmetry of the problem. This ensures that the electric field is perpendicular to the surface and constant in magnitude at every point on the cylinder.
2. Write the expression for the electric flux Φ_E, which is the electric field E times the cylindrical surface area (A). The area A of a cylinder can be calculated as 2πrl, where r is the radius of the cylindrical surface, and l is its length.
3. According to Gauss's theorem, the electric flux through this closed surface is equal to the charge enclosed (Q_enc) divided by ε_0.
So the electric field E due to a line charge with linear charge density λ (charge per unit length) is:
Φ_E = E · 2πrl = λl/ε_0
Here, l cancels out from both sides:
E · 2πr = λ/ε_0
Therefore, the electric field intensity E is given by:
E = λ / (2πε_0r)
c. i. The negatively charged oil drop is stationary because the electric force acting upwards on the drop is balancing the gravitational force pulling it downwards. The electric force is due to the electric field between the charged metal plates, and the gravitational force is due to the Earth's gravity. When the two forces are equal in magnitude and opposite in direction, the net force on the drop is zero, making the drop stationary.
ii. To find the charge (q) on the oil drop, set the gravitational force (mg) equal to the electric force (qE):
mg = qE
where m is the mass of the oil drop, g is the acceleration due to gravity, and E is the electric field strength. The electric field E between two plates with a potential difference (V) and distance (d) apart is given by:
E = V/d
Now plug in the values to find q:
q = (mg) / E = (mg) / (V/d)
Given values:
mass of the oil drop, m = 5×10^-15 kg,
acceleration due to gravity, g = 9.81 m/s^2,
potential difference, V = 1000 V,
distance between plates, d = 25 mm = 25×10^-3 m.
Using the above values, we get:
q = (5×10^-15 kg × 9.81 m/s^2) / (1000 V / 25×10^-3 m)
Calculating this gives us the charge q on the oil drop.
