An Interesting Quiz On Calculating Total Capacitance

Explanation
The correct answer is C1+C2+C3. In a parallel connection, the total capacitance is equal to the sum of the individual capacitances. Therefore, the total capacitance of the connection is given by the formula C1+C2+C3.

Explanation
When the separation between the plates of a capacitor is halved, the capacitance (C) of the capacitor increases by a factor of four according to the formula C = εA/d, where ε is the permittivity of the material between the plates, A is the area of the plates, and d is the separation between them. As the capacitance increases, the force of attraction between the plates (F) also increases, since F is directly proportional to the capacitance. Therefore, the force of attraction between the plates becomes four times stronger when the separation between them is halved.

Explanation
When a dielectric is inserted between the plates of a parallel plate capacitor, the capacitance increases. This is because the dielectric material reduces the electric field between the plates, allowing more charge to be stored for a given potential difference. The presence of the dielectric increases the capacitance by a factor equal to the dielectric constant of the material. Therefore, the capacitance increases when a dielectric is inserted between the parallel plate capacitor.

Explanation
The stored energy of a capacitor can be calculated using the formula: E = 1/2 * C * V^2, where E is the energy, C is the capacitance, and V is the voltage. In this case, the capacitance is given as 5 microfarads (5 * 10^-6 F) and the voltage is 100V. Plugging these values into the formula, we get: E = 1/2 * 5 * 10^-6 * 100^2 = 2.5/100 Joules. Therefore, the stored energy of the capacitor is 2.5/100 Joules.

Explanation
The thickness of the plate does not affect the capacitance of the capacitor because capacitance is determined by the area of the plates, the separation between them, and the permittivity of the dielectric material. The thickness of the plate does not play a role in determining the capacitance value.

Explanation
When capacitors are arranged in parallel, the total capacitance is equal to the sum of the individual capacitances. In this case, since the three capacitors have a capacitance of 4F each, the total capacitance would be 4F + 4F + 4F = 12F. Therefore, the correct answer is 12F.

Explanation
When capacitors are connected in series, the total capacitance is given by the reciprocal of the sum of the reciprocals of the individual capacitances. In this case, since the capacitance of both capacitors is 2F, the reciprocal of the sum of the reciprocals is 1/2F + 1/2F = 1F. Therefore, the total capacitance of the connection is 1F.

Explanation
When the potential difference and capacitance of a capacitor are doubled, the stored energy in the capacitor increases by a factor of 8. This can be explained by the formula for the energy stored in a capacitor, which is given by U = (1/2) * C * ∆V^2. When both C and ∆V are doubled, the energy becomes U' = (1/2) * (2C) * (2∆V)^2 = 8 * (1/2) * C * ∆V^2 = 8U. Therefore, the stored energy in the capacitor is now 8 times the original energy.

Explanation
The relationship between the capacitance of the capacitor and relative permittivity is that they are directly proportional. This means that as the relative permittivity of the material between the capacitor plates increases, the capacitance also increases. Conversely, if the relative permittivity decreases, the capacitance decreases as well. This relationship is described by the equation C = ε₀εᵣA/d, where C is the capacitance, ε₀ is the permittivity of free space, εᵣ is the relative permittivity, A is the area of the capacitor plates, and d is the distance between the plates.

Explanation
The capacitance of a capacitor opposes the sudden change in voltage. Capacitance is a measure of a capacitor's ability to store charge, and it is directly related to the amount of charge that can be stored per unit of voltage. When the voltage across a capacitor changes suddenly, the capacitance resists this change by either absorbing or releasing charge, thereby maintaining a more constant voltage. This property makes capacitors useful in smoothing out voltage fluctuations and stabilizing electrical circuits.