The purpose of this article is to learn the **interaction between an electric charge and an electric field**. We simply need to find the f**orces acting on the charged body** under the influence of the **electric field**.

To solve this question, we need to understand the **mathematical forms** of **electric field** and the **force acting on a charge** in an electric field.

The **force acting on two charges** due to their interaction is mathematically given by the following **formula**:

\[ F \ = \ \dfrac{ k \times Q \times q }{ r^{ 2 } }\]

The **electric field strength** of a charged body **at a distance** of $ r $ is given by the following mathematical **formula**:

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

**Expert Answer**

To **find the strength of the field at the point**, $(x,y)=(-5\:cm,-5\:cm)$.

To find the **distance, use the following formula:**

\[ r \ = \ \sqrt{ x^{ 2 } + y^{ 2 } } \]

\[ r \ = \ \sqrt{ ( -0.05 )^{ 2 } + ( -0.05 )^{ 2 } } \]

\[r \ = \ 0.071 \ m \]

The **distance** is $ r \ = \ 0.071 \ m $.

To find the **electric field strength** at the above point:

\[ E \ = \ \dfrac{ kq }{ r^{ 2 } } \]

**Plug values** of $ k $, $ q $ and $ r $.

\[ E \ = \ \dfrac{ ( 9 \times 10^{ 9 } ) ( 9 \times 10^{ -9 } ) }{ ( 0.071 )^{ 2 } } \]

\[ E \ = \ 1.8 \times 10^{ 4 } \dfrac{ N }{ C } \]

The **electric field strength** is $ E \ = \ 1.8 \times 10^{ 4 } \dfrac{ N }{ C } $.

**Numerical Result**

**E****lectric field strength at position** $ ( x, y ) \ = \ ( -5 \ cm, -5 \ cm ) $ is $ E \ = \ 1.8 \times 10^{ 4 } \dfrac{ N }{ C } $.

**Example**

**A $ +20 \ nC $ charge is located at the origin. What is the strength of the electric field at the position $ ( x, y ) = ( −6.0 \ cm, −6.0 \ cm ) $?**

**Solution**

To **find the strength of the field at the point**, $ ( x, y ) \ = \ ( -6 \ cm, -6 \ cm ) $.

To find the **distance, use the following formula:**

\[ r \ = \ \sqrt{ x^{ 2 } + y^{ 2 } } \]

\[ r \ = \ \sqrt{ ( -0.06 )^{ 2 } + ( -0.06 )^{ 2 } } \]

\[r \ = \ 0.0848 \ m \]

The **distance** is $ r \ = \ 0.0848 \ m $.

To find the **electric field strength** at the above point:

\[ E \ = \ \dfrac{ kq }{ r^{ 2 } } \]

**Plug values** of $ k $, $ q $ and $ r $.

\[ E \ = \ \dfrac{ ( 9 \times 10^{ 9 } )( 20 \times 10^{ -9 } ) }{ ( 0.0848 )^{ 2 } } \]

\[ E \ = \ 2.5 \times 10^{ 4 } \dfrac{ N }{ C } \]

**E****lectric field strength** is $ E \ = \ 2.5 \times 10^{ 4 } \dfrac{ N }{ C } $.

**Electric field strength at position** $ ( x, y ) \ = \ ( -6 \ cm, -6 \ cm ) $ is $ E \ = \ 2.5 \times 10^{ 4 } \dfrac{ N }{ C } $.