Electrostatic Potential And Capacitance

Connecting two capacitors in series and one capacitor in parallel allows for the effective capacitance to be calculated using the formula 1/Ceff = 1/C1 + 1/C2 + 1/C3. In this case, if two capacitors are connected in series, their effective capacitance would be 2 µF. Then, by connecting the third capacitor in parallel to the series combination, the effective capacitance would be increased to 6 µF, as desired.

When capacitors are connected in series, the reciprocal of the equivalent capacitance is equal to the sum of the reciprocals of the individual capacitances. So, for the given capacitors of capacitance 4µF, 6µF, and 12µF, the equivalent capacitance in series would be 1/(1/4 + 1/6 + 1/12) = 1/((3+2+1)/12) = 12/6 = 2µF.

When capacitors are connected in parallel, the equivalent capacitance is the sum of the individual capacitances. So, for the given capacitors, the equivalent capacitance in parallel would be 4µF + 6µF + 12µF = 22µF.

Therefore, the ratio of the equivalent capacitance in series to the equivalent capacitance in parallel is 2µF : 22µF, which simplifies to 1 : 11.

When the plates of the charged capacitor are connected by a conducting wire, the stored electrical energy in the capacitor is discharged. This discharge causes a flow of current through the wire, which leads to the generation of heat. The amount of heat produced can be calculated using the formula: Heat = 0.5 * C * V^2, where C is the capacitance and V is the voltage. In this case, the capacitance is given as 2µF and the voltage is 100 volts. Plugging these values into the formula gives us a heat of 0.01 J. Therefore, the correct answer is 0.01 J.

When capacitors are connected in series, the same amount of charge passes through each capacitor. This is because the capacitors are connected in a series circuit, so the current flowing through them is the same. Since the charge on a capacitor is directly proportional to the potential difference across it, and the potential difference is the same for capacitors connected in series, it follows that the charges on the capacitors will be equal. Therefore, option 1 is correct.

The work done in carrying a charge from infinity to a particular point in an electrostatic field is equal to the change in potential energy of the charge. Since the work done is positive (100 joules), it means that the potential energy of the charge has increased. In an electrostatic field, the potential decreases as we move away from the source of the field. Therefore, the potential at the particular point must be lower than infinity. The only option that satisfies this condition is -20 V, which indicates a potential 20 volts lower than infinity.

The potential energy between two charges can be calculated using the formula U = k * (q1 * q2) / r, where U is the potential energy, k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between them. In this case, the charges are negative and equal to -2 μC each, and the distance is 1m. Plugging these values into the formula, we get U = (8.99 * 10^9 N m²/C²) * (-2 μC * -2 μC) / 1m = 0.036 J.

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The time it takes to raise the potential of a capacitor can be calculated using the formula: time = charge/capacitance. In this case, the charge is given as 50 µC/sec and the capacitance is 800 µF. By substituting these values into the formula, we get time = 50 µC/sec / 800 µF = 0.0625 sec. However, the question asks for the time in seconds, so we need to convert the time to seconds by multiplying it by 1000. Therefore, the correct answer is 0.0625 sec * 1000 = 62.5 s.

The dielectric constant of a material is a measure of its ability to store electrical energy in an electric field. The capacitance of a parallel plate capacitor is directly proportional to the dielectric constant of the material between the plates. In this case, the capacitance of the capacitor increases from 50 pf to 105 pf when immersed in oil. By using the formula for capacitance, C = (k * ε₀ * A) / d, where k is the dielectric constant, ε₀ is the permittivity of free space, A is the area of the plates, and d is the distance between the plates, we can find that the dielectric constant of the oil is 2.1.

When capacitors are connected in parallel, the voltage across each capacitor is the same. The energy stored in a capacitor is given by the formula 1/2 * C * V^2, where C is the capacitance and V is the voltage. Since all the capacitors are connected in parallel to the same voltage source, the voltage V is constant for all capacitors. Therefore, the energy stored in each capacitor is equal to C * V. Since there are n capacitors in parallel, the total energy stored is equal to n * C * V, which can be simplified to nC.

The total capacitance of a parallel plate capacitor made by stocking n equally spaced plates connected alternately can be found by multiplying the number of plates (n) minus 1, with the capacitance between any two plates (x). This is because each plate is connected to two neighboring plates, except for the first and last plates which are only connected to one neighboring plate. Therefore, the total capacitance is equal to (n - 1) times x.

The potential at the center of the sphere is 80 volts because the potential at any point inside a hollow metal sphere is constant and equal to the potential on its surface. Since the potential on the surface is given as 80 volts, the potential at the center will also be 80 volts.

When air is replaced by a dielectric medium of constant K, the maximum force of attraction between two charges separated by a distance becomes K times. This is because the presence of a dielectric medium increases the electric field between the charges, which in turn increases the force of attraction between them. The constant K, also known as the dielectric constant, quantifies the extent to which the medium can increase the electric field. Therefore, the force of attraction between the charges becomes K times greater when air is replaced by a dielectric medium.

The correct answer is "Varies as 1/r, where r is the distance from the axis." In a charged cylindrical capacitor, the electric field magnitude decreases as the distance from the axis increases. This is because the electric field lines spread out as they move away from the axis, resulting in a decrease in field strength. The relationship between the electric field magnitude and the distance from the axis is inversely proportional, following the equation E ∝ 1/r.

The equivalent capacity of the combination C1 = C2 = C3 shown in the figure is 2C. This means that when these capacitors are connected in parallel, the total capacity is double the value of each individual capacitor.