# A +9 nC charge is located at the origin. What is the strength of the electric field at the position (x,y)=(−5.0 cm,−5.0 cm)

The purpose of this article is to learn the interaction between an electric charge and an electric field. We simply need to find the forces 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:

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

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

Electric 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 $r$.

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

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

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

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