The** aim of this question** is to understand the basic working principle of the **electrostatic precipitator** by applying the key concepts of **static electricity** including **electric field, electric potential, electrostatic force, etc.**

**Electrostatic precipitators** are used to remove** undesired particles** (especially **pollutants**) from smoke or **effluent gases**. They are used mostly in **coal-powered power plants** and **grain processing plants**. The simplest precipitator is a **vertically stacked hollow metallic cylinder** containing a **thin metallic wire** insulated from the outer cylindrical shell.

A** potential difference** is applied across the central wire and the cylindrical body that creates a **strong electrostatic field**. When the soot is passed through this cylinder, it **ionizes the air** and its constituent particles. The heavy metallic particles are attracted towards the central wire and hence the **air is cleaned.**

## Expert Answer

For an **electrostatic precipitator**, the magnitude of the **electric field** can be calculated by using the following equation:

\[ E \ = \ \dfrac{ V_{ ab } }{ ln( \frac{ b }{ a } ) } \times \dfrac{ 1 }{ r } \]

**Given that:**

\[ V_{ ab } \ = \ 50 \ kV \ = \ 50000 \ V \]

\[ b \ = \ 14 \ cm \ = \ 0.140 \ m \]

\[ a \ = \ 90 \ \mu m \ = \ 90 \times 10^{ -6 } \ m \]

\[ r \ = \ \dfrac{ 0.140 }{ 2 } \ m \ = \ 0.07 \ m \]

**Substituting the given values in the above equation:**

\[ E \ = \ \dfrac{ 50000 }{ ln( \frac{ 0.140 }{ 90 \times 10^{ -6 } } ) } \times \dfrac{ 1 }{ 0.070 } \]

\[ E \ = \ \dfrac{ 50000 }{ ln( 1555.56 ) \times 0.070 } \]

\[ E \ = \ \dfrac{ 50000 }{ 7.35 \times 0.070 } \]

\[ E \ = \ \dfrac{ 50000 }{ 0.51 } \]

\[ E \ = \ 98039.22\]

\[ E \ = \ 9.80 \times 10^{ 4 } \ V/m \]

## Numerical Result

\[ E \ = \ 9.80 \times 10^{ 4 } \ V/m \]

## Example

What will be the** electrostatic force** if we **half the applied potential difference?**

**Recall:**

\[ E \ = \ \dfrac{ V_{ ab } }{ ln( \frac{ b }{ a } ) } \times \dfrac{ 1 }{ r } \]

**Given that:**

\[ V_{ ab } \ = \ 25 \ kV \ = \ 25000 \ V \]

\[ b \ = \ 14 \ cm \ = \ 0.140 \ m \]

\[ a \ = \ 90 \ \mu m \ = \ 90 \times 10^{ -6 } \ m \]

\[ r \ = \ \dfrac{ 0.140 }{ 2 } \ m \ = \ 0.07 \ m \]

**Substituting the given values in the above equation:**

\[ E \ = \ \dfrac{ 25000 }{ ln( \frac{ 0.140 }{ 90 \times 10^{ -6 } } ) } \times \dfrac{ 1 }{ 0.070 } \]

\[ E \ = \ \dfrac{ 50000 }{ ln( 1555.56 ) \times 0.070 } \]

\[ E \ = \ \dfrac{ 50000 }{ 7.35 \times 0.070 } \]

\[ E \ = \ \dfrac{ 25000 }{ 0.51 } \]

\[ E \ = \ 49019.61 \]

\[ E \ = \ 4.90 \times 10^{ 4 } \ V/m \]