Electrical Engineering All Engineering Career

The concept of electric field lines and the visualization of electric field as divided into tubes of force containing a fixed number of lines of force was suggested by Michael Faraday. So, the correct answer is:

Faraday

For closer tolerances, mica capacitors are usually preferred. Mica capacitors can achieve tighter tolerances compared to paper or ceramic capacitors. So, the correct option is:

Mica capacitor

In the scenario described, where a capacitor becomes hot and ultimately explodes due to wrong connections, it's most likely an electrolytic capacitor. Electrolytic capacitors are polarized, meaning they have a positive and a negative terminal, and connecting them with reverse polarity can lead to catastrophic failure, including overheating and explosion.

So, the correct option is:

Electrolytic capacitor

A gang condenser is a variable capacitor. It consists of multiple capacitor sections mechanically connected together, allowing for simultaneous adjustment of capacitance across all sections. This design is often used in tuning circuits and radio frequency applications.

So, the correct option is:

variable capacitor

Capacitance increases with an increase in plate area and a decrease in the distance between the plates. This relationship is described by the formula for capacitance:

\[ C = \dfrac{{\varepsilon_0 \cdot \varepsilon_r \cdot A}}{{d}} \]

Where:
- \( C \) is the capacitance.
- \( \varepsilon_0 \) is the vacuum permittivity.
- \( \varepsilon_r \) is the relative permittivity (dielectric constant) of the material between the plates.
- \( A \) is the area of overlap of the plates.
- \( d \) is the separation distance between the plates.

Increasing the plate area (\( A \)) or decreasing the separation distance between the plates (\( d \)) both lead to an increase in capacitance.

So, the correct option is:

increase in plate area and decrease in distance between the plates

When the separation between the plates of a parallel plate capacitor is increased while keeping the voltage source constant, the capacitance \( C \) decreases according to the formula:

\[ C = \dfrac{{\varepsilon_0 \cdot \varepsilon_r \cdot A}}{{d}} \]

Where:
- \( \varepsilon_0 \) is the vacuum permittivity.
- \( \varepsilon_r \) is the relative permittivity (dielectric constant) of the material between the plates.
- \( A \) is the area of overlap of the plates.
- \( d \) is the separation distance between the plates.

As the capacitance decreases, the charge \( Q \) on the capacitor remains constant because it's determined by the voltage source and the capacitance: \( Q = CV \). Therefore, if \( C \) decreases, \( V \) must increase to keep \( Q \) constant.

So, the correct option is:

voltage across the plates increases

A dielectric material must be an insulator. Dielectric materials are characterized by their ability to support an electric field with minimal flow of electric current. They are used in capacitors to separate the plates and increase the capacitance by reducing the electric field strength between them.

So, the correct option is:

insulator

For making a capacitor, it is better to select a dielectric material with a high permittivity (also known as a high relative permittivity or a high dielectric constant). 

A dielectric material with a high permittivity allows for a greater charge storage capacity in the capacitor compared to vacuum or air, as it reduces the electric field between the plates for a given voltage, thereby increasing the capacitance. 

So, the correct option is:

high permittivity

Electrolytic capacitors must be used in circuits with the end marked positive always at a more positive potential than its negative end. Thus, they are normally used only where dc or pulsating dc voltages are present.

The capacitance of a capacitor is not affected by the thickness of the plates. The capacitance \( C \) of a parallel plate capacitor is determined by the formula:

\[ C = \dfrac{{\varepsilon_0 \cdot \varepsilon_r \cdot A}}{{d}} \]

Where:
- \( \varepsilon_0 \) is the vacuum permittivity.
- \( \varepsilon_r \) is the relative permittivity (dielectric constant) of the material between the plates.
- \( A \) is the area of overlap of the plates.
- \( d \) is the separation distance between the plates.

As you can see from the formula, the capacitance depends on the area of the plates (\( A \)) and the distance between them (\( d \)), but it's not affected by the thickness of the plates.

So, the correct option is:

thickness of plates