Electrostatic Potential And Capacitance

The given circuit has resistors arranged in series and parallel. To find the equivalent resistance, we need to simplify the circuit step by step. 1. First, the resistors 6Ω and 4Ω are in parallel. The equivalent resistance of two resistors in parallel is given by: 2. Next, the resistors 2Ω and 2.4Ω are in series. The total resistance is: 3. Finally, the resistors 3Ω, 4.4Ω, and 3Ω are in series. The total equivalent resistance is: However, since the problem states that the potential at A is greater than the potential at B, we need to look at the configuration more closely. After applying the simplifications correctly, the final equivalent resistance of the circuit across AB is .

The correct option is (E) :

Kirchhoff’s first law (junction rule) states that the sum of currents entering a junction is equal to the sum of currents leaving the junction. This is based on the principle of conservation of charge. Kirchhoff’s second law (loop rule) states that the sum of the potential differences (voltages) around any closed loop is zero. This law is based on the conservation of energy in the circuit. Therefore, the conservation of charge and energy is directly related to Kirchhoff’s junction rule and loop rule.

The correct option is (C) : Kirchhoff's junction rule and loop rule

We are given a combination of capacitors, and we need to calculate the effective capacitance between points A and B. Let's solve step by step. 1. First, consider the two capacitors of F each in series: 2. Now, this result is in parallel with the F capacitor: 3. Next, the two F capacitors in series again: 4. Finally, the result is in series with : Thus, the effective capacitance between A and B is .

The correct option is (B) :

Given parameters:

• Charges:
• Separation:
• Electric field:
• Angle:

Dipole moment calculation:

Torque on dipole:

Potential energy:

Thus, the correct option is (A): and .