This question aims at developing the understanding of the **electric field** and **potential gradient** around point charges.

Whenever** two charges** are placed in each other’s **vicinity**, they **exert force** on each other called the **C****oulomb’s electrostatic force, **which is mathematically defined as:

\[ F \ = \ k \dfrac{ q_1 q_2 }{ r^2 } \]

Where $ q_1 $ and $ q_2 $ are the **charges placed at a distance** $ r $ from each other.

This **force is due to the electric field** that exists between these two charges. The **electric field of a point charge** at a distance $ r $ is defined as:

\[ E \ = \ k \dfrac{ q }{ r^2 } \]

The **electric potential difference** at a point in an electric field is defined mathematically as:

\[ V_2 – V_1 \ = \ – E r \]

## Expert Answer

Let us **assume that** $ q_1 $ is placed at the origin and $ q_1 $ is placed at the $ a $ mark along the x-axis. Also, let $ x $ be the **distance at which the electric field is zero**.

Given:

\[ x \ =\ 15 \ cm \]

And the **total electric field: **

\[ E \ = \ E_1 \ + \ E_2 \]

Where $ E_1 $ and $ E_2 $ are the **electric fields due to each** of the $ q_1 $ and $ q_2 $ charges respectively. Using the **formula for electric field**:

\[ E \ = \ k \dfrac{ q }{ r^2 } \]

For $ q_1 $:

\[ E_1 \ = \ k \dfrac{ q_1 }{ x^2 } \]

For $ q_2 $:

\[ E_2 \ = \ – k \dfrac{ q_2 }{ ( 15 – x )^2 } \]

The **negative sign** shows that the **direction is opposite** to the x-axis. **Substituting these values** in the total electric field equation:

\[ E \ = \ k \dfrac{ q_1 }{ x^2 } \ – \ k \dfrac{ q_2 }{ ( 15 – x )^2 } \]

At the point $ x $, the **total electric field has to be zero**, so:

\[ 0 \ = \ k \dfrac{ q_1 }{ x^2 } \ – \ k \dfrac{ q_2 }{ ( 15 – x )^2 } \]

\[ k \dfrac{ q_2 }{ ( 15 – x )^2 } \ = \ k \dfrac{ q_1 }{ x^2 } \]

\[ \dfrac{ q_2 }{ ( 15 – x )^2 } \ = \ \dfrac{ q_1 }{ x^2 } \]

\[ q_2 x^2 \ = \ q_1 ( 15 – x )^2 \]

\[ q_2 x^2 \ = \ q_1 ( 15^2 – 2( 15 )( x ) + x^2 ) \]

\[ q_2 x^2 \ = \ q_1 ( 225 – 30 x + x^2 ) \]

\[ q_2 x^2 \ = \ 225 q_1 – 30 x q_1 + x^2 q_1 \]

\[ 0 \ = \ 225 q_1 – 30 x q_1 + x^2 q_1 – x^2 q_2 \]

\[ 0 \ = \ 225 q_1 + (- 30 q_1 ) x + ( q_1 – q_2 ) x^2 \]

\[ 225 q_1 + (- 30 q_1 ) x + ( q_1 – q_2 ) x^2 \ = \ 0 \]

**Substituting values:**

\[ 225 \times 10 + (- 30 \times 10 ) x + ( 10 – 20 ) x^2 \ = \ 0 \]

\[ 2250 + (- 300 ) x + ( – 10 ) x^2 \ = \ 0 \]

**Using the quadratic roots formula:**

\[ x \ =\ \dfrac{ – ( -300 ) \pm \sqrt{ (-300)^2 – 4 ( 2250 )( -10 ) } }{ 2 ( -10 ) } \]

\[ x \ =\ \dfrac{ 300 \pm \sqrt{ 90000 + 90000 } }{ -20 } \]

\[ x \ =\ – \dfrac{ 300 \pm \sqrt{ 180000 } }{ 20 } \]

\[ x \ =\ – \dfrac{ 300 \pm 424.26 }{ 20 } \]

\[ x \ =\ – \dfrac{ 300 + 424.26 }{ 20 }, \ – \dfrac{ 300 – 424.26 }{ 20 } \]

\[ x \ =\ – \dfrac{ 724.26 }{ 20 }, \ – \dfrac{ – 124.26 }{ 20 } \]

\[ x \ =\ – 36.213 \ cm , \ 6.21 \ cm \]

## Numerical Result

\[ x \ =\ – 36.213 \ cm , \ 6.21 \ cm \]

## Example

Calculate the **magnitude of the electric field at a distance of 5 cm** from a 10 nC charge.

\[ E \ = \ k \dfrac{ q_1 }{ x^2 } \ – \ k \dfrac{ q_2 }{ ( 0.15 – x )^2 } \]

**Substituting values:**

\[ E \ = \ 9 \times 10^9 \dfrac{ 10 \times 10^{-9} }{ ( 0.05 )^2 } \ – \ 9 \times 10^9 \dfrac{ 20 \times 10^{-9} }{ ( 0.15 – 0.05 )^2 } \]

\[ E \ = \ \dfrac{ 90 }{ 0.0025 } \ – \ \dfrac{ 180 }{ 0.01 } \]

\[ E \ = \ 36000 \ – \ 18000 \]

\[ E \ = \ 18000 \ N/C \]