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Is there a point between a 10 nC charge and a 20 nC charge at which the electric field is zero? What is the electric potential at this point if both charges are separated by 15 cm?

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 Coulomb’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:

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

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

\[ E \ = \ 36000 \ – \ 18000 \]

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

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