Electrostatics

Step 1: For an ideal gas, PV = nRT
Step 2: At constant temperature, PV = constant
Step 3: Hence, a plot of PV versus V is a horizontal straight line.

Step 1: Capacitance of parallel plate capacitor, C=(₀ A)/(d)
Step 2: Substituting values, 1=8.85×10⁻¹²A10⁻³
Step 3: A≈1.13×10⁹ m²≈18×10⁸ m²

Step 1: Potential gradient k 1/L.
Step 2: New length, L'=+()/(2)=(3)/(2)
Step 3: Balance length varies inversely with length, '=()/(3)×()/(L')=()/(6)

For a closed current loop in a uniform magnetic field, net magnetic force is zero though torque may act.

Step 1: Shunt resistance, S=(Ig Rg)/(I-Ig)
Step 2: Substituting values, S=(1×0.81)/(10-1)=0.09≈0.03 Ω

Lenz’s law ensures that the induced current always opposes the cause producing it, thereby conserving energy.

Step 1: Sharpness of resonance depends on damping.
Step 2: Smaller damping results in a higher quality factor Q.
Step 3: Higher Q produces a sharper resonance peak.

Step 1: The positive charge at the support repels the positively charged bob.
Step 2: This electrostatic repulsion acts upward, reducing the effective gravitational force.
Step 3: Hence gₑₓtₑff < g, and T = 2π√Lgₑₓtₑff increases.

Step 1: For a closed organ pipe, f (1)/(L)
Step 2: Ratio of frequencies, (f₁)/(f₂)=(L₂)/(L₁)=(25)/(24)
Step 3: Given beat frequency, |f₁-f₂|=6
Step 4: Solving gives, f₁=120 Hz, f₂=124 Hz

An electron in a uniform electric field experiences constant acceleration perpendicular to its initial velocity, resulting in a parabolic trajectory.

We are asked to find the electric field at a point due to a uniformly charged spherical shell. The solution involves understanding the application of Gauss's Law, which is a fundamental concept in electrostatics.
1. Gauss's Law states that the electric flux through a closed surface is proportional to the charge enclosed by that surface: Where: - is the electric field, - is an infinitesimal area element, - is the enclosed charge, - is the permittivity of free space.
2. For a spherical shell, the symmetry of the charge distribution means that the electric field inside the shell must be uniform and radially symmetric. - At any point inside the spherical shell, the contribution to the electric field from all parts of the shell cancels out due to symmetry. - This is because the field due to a charge element at one part of the shell is canceled by the field due to a charge element at the opposite side of the shell. Thus, the electric field inside the shell is zero.
3. Therefore, the electric field at a point inside the shell is zero. This result is independent of the radius of the shell or the charge enclosed by the shell.
4. Outside the shell (for ), the electric field behaves as though the entire charge is concentrated at the center, but the question asks for the field at a point inside the shell. Thus, option (1), (3), and (4) are incorrect.
Option (2) is correct because the electric field inside the shell is zero.

The electric field is defined as force per unit charge: Force is in Newtons (N), charge is in Coulombs (C), so: Thus, option (3) is correct.

Step 1: Define electric charge
Electric charge is a fundamental property of matter that causes it to experience a force in an electric field. The SI unit for electric charge is defined in the context of electric current.

Step 2: Analyze each option
- Coulomb (C): The Coulomb is the SI unit of electric charge. It is defined as the charge transported by a constant current of 1 ampere in 1 second: - Ampere (A): The Ampere is the SI unit of electric current, measuring the flow of charge per unit time: - Volt (V): The Volt is the SI unit of electric potential difference or voltage, defined as: - Joule (J): The Joule is the SI unit of energy or work, defined as:
Step 3: Conclusion
Only the Coulomb is the unit of electric charge.

Given the formula for the electric field at a point on the axis of a uniformly charged ring: Substitute : Simplifying: Thus, the electric field is:

The force acting on the particle due to the electric field is: Where is the charge and is the electric field strength. The weight of the particle is given by: Since the particle is in equilibrium, the electric force balances the weight: The charge of one electron is . The number of electrons removed is: Thus, the correct number of electrons removed is .

Step 1: Identify the initial conditions. The first capacitor ( ) is charged to a potential difference of 100 V. The second capacitor ( ) is uncharged, so its initial charge is 0.
Step 2: Calculate the initial charge on the first capacitor. The charge on a capacitor is given by :
Step 3: Determine the total capacitance when connected in parallel. Since the capacitors are connected together, they share the same final potential difference, and their capacitances add up:
Step 4: Apply the conservation of charge. When connected, the charge from the first capacitor redistributes between both capacitors. The total initial charge is conserved: where is the final potential difference. Substitute the values:
Step 5: Solve for the final potential difference:
Step 6: Verify the result. The final potential difference should be less than the initial 100 V due to charge sharing, and the calculated value 71.43 V is consistent with the options.