Electrostatics MCQ Questions & Answers | Electrical Engineering

Option A: low permittivity

A dielectric with low permittivity would result in a capacitor with lower capacitance. This is not desirable for most applications where higher capacitance is needed.

Option B: high permittivity

A dielectric with high permittivity increases the capacitance of a capacitor. Higher permittivity means the dielectric material can store more electric energy for the same electric field, making it a better choice for capacitors.

Option C: permittivity same as that of air

Using a dielectric with the same permittivity as air would not provide any advantage over using air itself. It would not enhance the capacitance of the capacitor significantly.

Option D: permittivity slightly more than that of air

While a dielectric with permittivity slightly more than that of air would increase the capacitance somewhat, it would not be as effective as a dielectric with high permittivity.

Conclusion:

To make a capacitor with higher capacitance, it is better to select a dielectric with high permittivity, making Option B the correct answer.

Option A: Mica
Mica capacitors are known for their high breakdown voltage capability. They can withstand higher voltages compared to many other types of capacitors.

Option B: Paper
Paper capacitors typically have moderate breakdown voltage ratings. They are used in applications where moderate voltage levels are encountered.

Option C: Ceramic
Ceramic capacitors usually have breakdown voltages that vary widely depending on their construction. They can be designed for various voltage ratings, from low to high.

Option D: Electrolytic
Electrolytic capacitors generally have the lowest breakdown voltage among the options listed. They are designed for applications where moderate to low voltage levels are involved. Exceeding their breakdown voltage can cause permanent damage.

Conclusion:
Among the given options, electrolytic capacitors typically have the least value of breakdown voltage, making Option D the correct answer.

Here's the explanation for the capacitance problem:

The question is about finding the capacitance of a capacitor.

First, remember the basic formula that connects charge (Q), voltage (V), and capacitance (C):
Q = C * V
where:
Q is the charge stored in the capacitor (in Coulombs).
C is the capacitance (in Farads).
V is the voltage across the capacitor (in Volts).

We are given:
Q = 2000 μC (which is 2000 * 10^-6 Coulombs)
V = 200 V

We need to find C.

Rearrange the formula to solve for C:
C = Q / V

Substitute the given values:
C = (2000 * 10^-6) / 200
C = 10 * 10^-6 F
C = 10 μF (microFarads)

So, the correct answer is Option B: 10 uF.