Electrostatic Potential And Capacitance

Electric field intensity at the centre of the disc. (given)

Electric field along the axis at any distance x from the centre of the disc

From question, x=R (radius of disc)

% reduction in the value of electric field

Hence, The correct answer is option (A): reduces by 70.7%

Let's determine the ratio of the equivalent capacities when two equal capacitors are connected in series and in parallel.

1. Consider the capacitance of each capacitor as .
2. Capacitors in Series:
   When capacitors are connected in series, the reciprocal of the total capacitance ( ) is the sum of the reciprocals of the individual capacitances:
   Simplifying gives:
3. Capacitors in Parallel:
   When capacitors are connected in parallel, the total capacitance ( ) is the sum of the individual capacitances:
4. Finding the Ratio:
   The ratio of the equivalent capacities in the two cases is given by:

The ratio of equivalent capacitances when connected in series to that when connected in parallel is 1:4.

Thus, the correct answer is: 1:04.

The correct answer is (D) :
Electric field inside material of conductor is zero.

The correct answer is (D) :

For a uniformly charged non-conducting sphere, the potential inside the sphere (at a distance from the center) is given by:

Here:

• is the total charge on the sphere.
• is the radius of the sphere.
• is the distance from the center (for the center, ).

At the center ( ):

At the surface ( ), the potential is:

From this, we see:

Therefore, the potential at the center is:

Answer

The potential at the center of the sphere is:

The capacitance of a parallel plate capacitor is given by:

where:

• is the permittivity of free space,
• is the area of the plates,
• is the separation between the plates.

When a dielectric of constant is introduced, the new capacitance becomes:

In the given problem:

• Initial capacitance
• The separation between the plates becomes twice ( ),
• A dielectric constant is introduced.

The new capacitance is:

Substitute and :

The problem states that . Equating:

Thus, the value of is 105.

The correct answer is 225.

The correct answer is 6.

Understanding the Problem

We are given three capacitors connected in parallel, with capacitances 5 μF, x μF, and 2 μF. The total charge stored is 100 μC when a potential difference of 10 V is applied. We need to find the value of x.

Solution

1. Use the Formula for Total Capacitance in Parallel:

When capacitors are connected in parallel, the total capacitance is the sum of the individual capacitances:

Given , , and , we have:

2. Use the Formula for the Charge Stored in a Capacitor:

The charge stored on the capacitors is given by the formula:

From the problem, we know that the total charge and . Substituting the values into the equation:

3. Solve for x:

Corrected Final Answer

Thus, the value of is .

Note: The original calculation had an error in summing the constant terms.

To find the potential difference between points P and Q due to a charge placed at the origin, we use the formula for electric potential due to a point charge:

where is the electric potential, is the Coulomb's constant, is the charge, and is the distance from the charge.

1. Calculate the distance from the charge at the origin to point P :
2. Determine the distance from the charge to point Q :
3. Since both points P and Q are equidistant from the origin, the potentials at P and Q are the same:
4. Thus, the potential difference between the two points P and Q is:

The correct answer is 0 V. The potential at two equidistant points from the same charge is equal, leading to a potential difference of zero.

The problem aims to determine the electric potential due to a system of charged wires. The fundamental principle used is that the electric potential is the negative line integral of the electric field.

1. Electric Potential Definition:
The electric potential is defined as the negative line integral of the electric field along a path :

2. Electric Field of a Single Charged Wire:
The electric field due to an infinitely long charged wire with linear charge density at a radial distance is given by:
where is Coulomb's constant.

3. Potential Due to a Single Wire:
Integrating the electric field to find the potential due to a single wire:
where is the constant of integration.

4. Potential Due to All Wires:
Assuming we have three wires located in such a way that the distances from the point of interest to each wire are given by , , and , the total potential is the sum of the potentials from each wire:
(Note: I've changed the sign here, as the solution in the prompt seems to have dropped the negative signs, or chosen a slightly different reference point for 0 potential.)

5. Simplification and Final Expression:
Combine the logarithmic terms:

6. Setting : If is constant (equal to ), then: Since C' is some other constant: Where is another constant.

Final Conclusion:
Therefore, the condition for constant potential is:

The given circuit consists of two logic gates:
1. The first gate is an AND gate.
2. The second gate is an OR gate. Let's evaluate the output for each pair of and :
- When and , the output of the AND gate is because both inputs are zero. The output of the OR gate is also , since the OR gate only outputs 1 when at least one input is 1. Thus, the final output is low (zero). - When and , the output of the AND gate is . The OR gate outputs , but since the AND gate's output is zero, the final output is still low.
- When and , the output of the AND gate is because the second input is zero. The OR gate outputs , but the final output will still be low.
- When and , the AND gate outputs , and the OR gate also outputs , resulting in a high output.
Thus, the output is low for the following combinations: - (A) - (B) - (C)
Final Answer: (1) (A), (B) and (C) only.

The potential due to a long charged wire is proportional to the logarithm of the distance from the wire. To find the equipotential surface, we sum the potentials from the three wires.
Step 1: For each wire, the potential depends on the perpendicular distance from the wire.
Step 2: The equipotential surface is where the total potential from the three wires is constant.
Step 3: After analyzing the expressions, we conclude that the correct relation is , which satisfies the condition for an equipotential surface.

The path difference is , which corresponds to a phase difference between the two light waves. Since the polarizer is at , it ensures that the intensity is maximized for this configuration. Therefore, the resulting intensity after is:

Final Answer: .

Assertion (A) is true because a choke coil is indeed used to limit the current in a circuit with fluorescent tubes, and its high inductance ensures that it behaves like a resistor in the AC circuit.
Reason (R) is false because the voltage reduction factor should be more accurately derived based on the reactance of the coil and the resistance, and the reason provided is incomplete and incorrect in this context. Thus, the correct answer is (4).