Forces Between Charges

Lets us express the force between two point charges in terms of their position vector.

Figure:2.a

Consider two point charges $q_{1}$and $q_{2}$are at P and Q respectively. The position vector of point P is = <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mrow><mi>O</mi><mi>P</mi></mrow><mo>&#x2192;</mo></mover><mo>=</mo><mover><msub><mi>r</mi><mn>1</mn></msub><mo>&#x2192;</mo></mover></math> and that of point Q is <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mrow><mi>O</mi><mi>Q</mi></mrow><mo>&#x2192;</mo></mover><mo>=</mo><mover><msub><mi>r</mi><mn>2</mn></msub><mo>&#x2192;</mo></mover></math> .

Let $r$ be the distance between these charges.

We have,

<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mover><mi>F</mi><mo>&#x2192;</mo></mover><mn>12</mn></msub><mo>=</mo><mfrac><mn>1</mn><mrow><mn>4</mn><msub><mi>&#x3C0;&#x3B5;</mi><mn>0</mn></msub></mrow></mfrac><mfrac><mrow><msub><mi>q</mi><mn>1</mn></msub><msub><mi>q</mi><mn>2</mn></msub></mrow><msup><mi>r</mi><mn>2</mn></msup></mfrac><msub><mover><mi>r</mi><mo>^</mo></mover><mn>21</mn></msub></math>

Where <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mover><mi>r</mi><mo>^</mo></mover><mn>21</mn></msub><mo>=</mo><mfrac><msub><mover><mi>r</mi><mo>&#x2192;</mo></mover><mn>21</mn></msub><mi>r</mi></mfrac></math> is the unit vector in direction from $q_{2}$ to $q_{1}$.

The above equation can be rewritten as,

<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mover><mi>F</mi><mo>&#x2192;</mo></mover><mn>12</mn></msub><mo>=</mo><mfrac><mn>1</mn><mrow><mn>4</mn><msub><mi>&#x3C0;&#x3B5;</mi><mn>0</mn></msub></mrow></mfrac><mfrac><mrow><msub><mi>q</mi><mn>1</mn></msub><msub><mi>q</mi><mn>2</mn></msub></mrow><msup><mi>r</mi><mn>3</mn></msup></mfrac><mi>r</mi><mo>&#xA0;</mo><msub><mover><mi>r</mi><mo>^</mo></mover><mn>21</mn></msub></math> ……………….. (1)

Using triangle law of vector addition, we have,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mrow><mi>O</mi><mi>Q</mi></mrow><mo>&#x2192;</mo></mover><mo>+</mo><mover><mrow><mi>Q</mi><mi>P</mi></mrow><mo>&#x2192;</mo></mover><mo>=</mo><mover><mrow><mi>O</mi><mi>P</mi></mrow><mo>&#x2192;</mo></mover><mspace linebreak="newline"/><mi>o</mi><mi>r</mi><mo>,</mo><mo>&#xA0;</mo><mover><mrow><mi>Q</mi><mi>P</mi></mrow><mo>&#x2192;</mo></mover><mo>=</mo><mover><mrow><mi>O</mi><mi>P</mi></mrow><mo>&#x2192;</mo></mover><mo>-</mo><mover><mrow><mi>O</mi><mi>Q</mi></mrow><mo>&#x2192;</mo></mover></math>

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>o</mi><mi>r</mi><mo>,</mo><mo>&#xA0;</mo><mi>r</mi><mo>&#xA0;</mo><msub><mover><mi>r</mi><mo>^</mo></mover><mn>21</mn></msub><mo>=</mo><mover><msub><mi>r</mi><mn>1</mn></msub><mo>&#x2192;</mo></mover><mo>-</mo><mover><msub><mi>r</mi><mn>2</mn></msub><mo>&#x2192;</mo></mover></math>

And <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>o</mi><mi>r</mi><mo>,</mo><mo>&#xA0;</mo><mi>r</mi><mo>&#xA0;</mo><mo>=</mo><mfenced open="|" close="|"><mrow><mover><msub><mi>r</mi><mn>1</mn></msub><mo>&#x2192;</mo></mover><mo>-</mo><mover><msub><mi>r</mi><mn>2</mn></msub><mo>&#x2192;</mo></mover></mrow></mfenced><mo>&#xA0;</mo></math>

Putting these values in equation (1),

<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mover><mi>F</mi><mo>&#x2192;</mo></mover><mn>12</mn></msub><mo>=</mo><mfrac><mn>1</mn><mrow><mn>4</mn><msub><mi>&#x3C0;&#x3B5;</mi><mn>0</mn></msub></mrow></mfrac><mfrac><mrow><msub><mi>q</mi><mn>1</mn></msub><msub><mi>q</mi><mn>2</mn></msub></mrow><msup><mfenced open="|" close="|"><mrow><mover><msub><mi>r</mi><mn>1</mn></msub><mo>&#x2192;</mo></mover><mo>-</mo><mover><msub><mi>r</mi><mn>2</mn></msub><mo>&#x2192;</mo></mover></mrow></mfenced><mn>3</mn></msup></mfrac><mfenced><mrow><mover><msub><mi>r</mi><mn>1</mn></msub><mo>&#x2192;</mo></mover><mo>-</mo><mover><msub><mi>r</mi><mn>2</mn></msub><mo>&#x2192;</mo></mover></mrow></mfenced></math>

Similarly,

<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mover><mi>F</mi><mo>&#x2192;</mo></mover><mn>21</mn></msub><mo>=</mo><mfrac><mn>1</mn><mrow><mn>4</mn><msub><mi>&#x3C0;&#x3B5;</mi><mn>0</mn></msub></mrow></mfrac><mfrac><mrow><msub><mi>q</mi><mn>1</mn></msub><msub><mi>q</mi><mn>2</mn></msub></mrow><msup><mfenced open="|" close="|"><mrow><mover><msub><mi>r</mi><mn>2</mn></msub><mo>&#x2192;</mo></mover><mo>-</mo><mover><msub><mi>r</mi><mn>1</mn></msub><mo>&#x2192;</mo></mover></mrow></mfenced><mn>3</mn></msup></mfrac><mfenced><mrow><mover><msub><mi>r</mi><mn>2</mn></msub><mo>&#x2192;</mo></mover><mo>-</mo><mover><msub><mi>r</mi><mn>1</mn></msub><mo>&#x2192;</mo></mover></mrow></mfenced></math>

Superposition:

The principle of superposition gives the method to find the force on a charge when the system consists of many charges. When multiple charges are interacting, the total force on a given charge is given by,

<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mover><mi>F</mi><mo>&#x2192;</mo></mover><mn>1</mn></msub><mo>=</mo><msub><mover><mi>F</mi><mo>&#x2192;</mo></mover><mn>12</mn></msub><mo>+</mo><msub><mover><mi>F</mi><mo>&#x2192;</mo></mover><mn>13</mn></msub><mo>+</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>+</mo><msub><mover><mi>F</mi><mo>&#x2192;</mo></mover><mrow><mn>1</mn><mi>n</mi></mrow></msub></math>

Considering all the charges as negative, the electrostatic forces between these charges are the force of repulsion as like charges repel each other.

principle of superposition of electric charges

Figure:2.b

Therefore, using Coulomb’s law,

<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mover><mi>F</mi><mo>&#x2192;</mo></mover><mn>12</mn></msub><mo>=</mo><mfrac><mrow><msub><mi>q</mi><mn>1</mn></msub><msub><mi>q</mi><mn>2</mn></msub><msub><mover><mi>r</mi><mo>^</mo></mover><mn>12</mn></msub></mrow><msubsup><mi>r</mi><mn>12</mn><mn>2</mn></msubsup></mfrac></math> ,

<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mover><mi>F</mi><mo>&#x2192;</mo></mover><mn>13</mn></msub><mo>=</mo><mfrac><mrow><msub><mi>q</mi><mn>1</mn></msub><msub><mi>q</mi><mn>3</mn></msub><msub><mover><mi>r</mi><mo>^</mo></mover><mn>13</mn></msub></mrow><msubsup><mi>r</mi><mn>13</mn><mn>2</mn></msubsup></mfrac><mo>,</mo><mo>&#xA0;</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>&#xA0;</mo><mo>,</mo><mo>&#xA0;</mo><msub><mover><mi>F</mi><mo>&#x2192;</mo></mover><mrow><mn>1</mn><mi>n</mi></mrow></msub><mo>=</mo><mfrac><mrow><msub><mi>q</mi><mn>1</mn></msub><msub><mi>q</mi><mi>n</mi></msub><msub><mover><mi>r</mi><mo>^</mo></mover><mrow><mn>1</mn><mi>n</mi></mrow></msub></mrow><msubsup><mi>r</mi><mrow><mn>1</mn><mi>n</mi></mrow><mn>2</mn></msubsup></mfrac></math>

Total force is given by,

An example:

Three point charges $Q_{1}=4\times 10^{-9}C=4 nC$, $Q_{2}=6\times 10^{-9}C=6 nC$, and $Q_{3}=-3\times 10^{-9}C=-3 nC$form a right angle triangle. The distances between them are shown below. Find the magnitude of electrostatic force on charge $Q_{1}$due to $Q_{2}$ and $Q_{3}$. Given, dielectric constant = $9\times 10^{9}Nm^{2}C^{-2}$

Figure:2.c

The magnitude of electrostatic force exerted by $Q_{2}$on $Q_{1}$ is given by,

$F_{2}=k\frac{4\times 10^{-9}\times 6\times 10^{-9}}{(5\times 10^{-2})^{2}}$

Or, $F_{2}=k\frac{4\times 10^{-9}\times 6\times 10^{-9}}{25\times 10^{-4}}$

Or, $F_{2}=k\frac{4\times 10^{-9}\times 6\times 10^{-9}}{25\times 10^{-4}}=0.96 k\times 10^{-14}$

Putting $k=9\times 10^{9}$, we have,

Or, $F_{2}=9\times 10^{9}\times 0.96 \times 10^{-14}=8.64\times 10^{-5}$N

The magnitude of electrostatic force exerted by $Q_{3}$ on $Q_{1}$ is given by,

$F_{3}=k\frac{4\times 10^{-9}\times (3\times 10^{-9})}{(3\times 10^{-2})^{2}}=9\times 10^{9} \times \frac{12\times 10^{-18}}{9\times 10^{-4}}= 12\times 10^{-5}$N

The magnitude of the resultant force is given by,

$F_{R}^{2}=F_{2}^{2}+F_{3}^{2}$

Or, $F_{R}^{2}=$$(8.64\times 10^{-5})^{2}+(12\times 10^{-5})^{2}$

Or, $F_{R}=\sqrt{(8.64\times 10^{-5})^{2}+(12\times 10^{-5})^{2}}$

Or, $F_{R}=\sqrt{(74.64\times 10^{-10})+(144\times 10^{-10})}$

Or, $F_{R}=\sqrt{(218.64\times 10^{-10})}$

Or, $F_{R}=1.479\times 10^{-4}$N.

Figure:2.d

$tan \theta =$$12\times 10^{-5}$/$8.64\times 10^{-5}$

Or, $tan \theta =$$1.39$

Or, $\theta =tan^{-1}1.39=54.26^{0}$ to the negative x-axis.

 

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