Electric Potential

Electrostatic potential at any point of an electric field is defined as potential energy per unit charge at that point.

Electric Potential Energy:

Let us consider a system of two point charges $q$ and $q^{‘}$ (positive test charge) separated by a distance $r$

Figure:7.a

Charge $q$ is fixed at point $P$ and is displaced from point R to S along a radial line PRS shown in the figure.

According to Coulomb’s law, the magnitude of the force on a positive test charge as is given by,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>F</mi><mo>=</mo><mfrac><mn>1</mn><mrow><mn>4</mn><msub><mi>πε</mi><mn>0</mn></msub></mrow></mfrac><mfrac><mrow><mi>q</mi><msup><mi>q</mi><mo>'</mo></msup></mrow><msup><mi>r</mi><mn>2</mn></msup></mfrac></math>

If q’ moves towards $S$ through a small displacement $dr$ then work done by this force is given by,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>d</mi><mi>W</mi><mo>=</mo><mi>F</mi><mo>.</mo><mi>d</mi><mi>r</mi><mo>=</mo><mfrac><mn>1</mn><mrow><mn>4</mn><msub><mi>πε</mi><mn>0</mn></msub></mrow></mfrac><mfrac><mrow><mi>q</mi><msup><mi>q</mi><mo>'</mo></msup></mrow><msup><mi>r</mi><mn>2</mn></msup></mfrac><mi>d</mi><mi>r</mi></math>

Total work done from point $R$ to $S$ i.e., from $r_{1}$ to $r_{2}$ is given by,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>W</mi><mo>=</mo><mfrac><mn>1</mn><mrow><mn>4</mn><msub><mi>πε</mi><mn>0</mn></msub></mrow></mfrac><msubsup><mo>∫</mo><msub><mi>r</mi><mn>1</mn></msub><msub><mi>r</mi><mn>2</mn></msub></msubsup><mfrac><mrow><mi>q</mi><msup><mi>q</mi><mo>'</mo></msup></mrow><msup><mi>r</mi><mn>2</mn></msup></mfrac><mi>d</mi><mi>r</mi><mspace linebreak="newline"/><mo> </mo><mo> </mo><mo> </mo><mo>=</mo><mo> </mo><mfrac><mrow><mi>q</mi><msup><mi>q</mi><mo>'</mo></msup></mrow><mrow><mn>4</mn><msub><mi>πε</mi><mn>0</mn></msub></mrow></mfrac><mfenced><mrow><mfrac><mn>1</mn><msub><mi>r</mi><mn>1</mn></msub></mfrac><mo>-</mo><mfrac><mn>1</mn><msub><mi>r</mi><mn>2</mn></msub></mfrac></mrow></mfenced></math>

It shows that work done on the test charge depends on end points not on the path taken.

The work done in moving the test charge $q’^{}$ from point $R$ to $S$ is equal to the change in potential energy in moving the test charge $q’^{}$ from point $R$ to $S$.

Thus,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>W</mi><mo>=</mo><msub><mi>U</mi><msub><mi>r</mi><mn>1</mn></msub></msub><mo>-</mo><msub><mi>U</mi><msub><mi>r</mi><mn>2</mn></msub></msub></math>

where

<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><msub><mi>U</mi><mi>r</mi></msub><mn>1</mn></msub><mo> </mo><mo>=</mo><mo> </mo><mfrac><mrow><mi>q</mi><msup><mi>q</mi><mo>'</mo></msup></mrow><mrow><mn>4</mn><msub><mi>πε</mi><mn>0</mn></msub><msub><mi>r</mi><mn>1</mn></msub></mrow></mfrac></math>

is the potential energy of the test charge $q’^{}$ when it is at point $R$

<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><msub><mi>U</mi><mi>r</mi></msub><mn>2</mn></msub><mo> </mo><mo>=</mo><mo> </mo><mfrac><mrow><mi>q</mi><msup><mi>q</mi><mo>'</mo></msup></mrow><mrow><mn>4</mn><msub><mi>πε</mi><mn>0</mn></msub><msub><mi>r</mi><mn>2</mn></msub></mrow></mfrac></math>

is the potential energy of the test charge $q’^{}$ when it is at point $S$.

Thus potential energy of the test charge $q^{‘}$ at any distance r from charge $q$ is given by

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>U</mi><mo> </mo><mo>=</mo><mo> </mo><mfrac><mrow><mi>q</mi><msup><mi>q</mi><mo>'</mo></msup></mrow><mrow><mn>4</mn><msub><mi>πε</mi><mn>0</mn></msub><mi>r</mi></mrow></mfrac></math>

Electric potential:

Electric potential $V=U/q^{‘}$.

It is a scalar quantity and its SI unit is Volt.

1 Volt = 1 JC$^{-1}$

We have,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>W</mi><mo>=</mo><msub><mi>U</mi><msub><mi>r</mi><mn>1</mn></msub></msub><mo>-</mo><msub><mi>U</mi><msub><mi>r</mi><mn>2</mn></msub></msub></math>

Now dividing both sides of the above equation by $q^{‘}$, we get,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mi>W</mi><msup><mi>q</mi><mo>'</mo></msup></mfrac><mo>=</mo><mfrac><msub><mi>U</mi><msub><mi>r</mi><mn>1</mn></msub></msub><mrow><mi>q</mi><mo>'</mo></mrow></mfrac><mo>-</mo><mfrac><msub><mi>U</mi><msub><mi>r</mi><mn>2</mn></msub></msub><msup><mi>q</mi><mo>'</mo></msup></mfrac><mo>=</mo><msub><msub><mi>V</mi><mi>r</mi></msub><mn>1</mn></msub><mo>-</mo><msub><msub><mi>V</mi><mi>r</mi></msub><mn>2</mn></msub></math>

Where $V_{r_{1}}$is the potential energy per unit charge at point $R$ and $V_{r_{2}}$is the potential energy per unit charge at point $S$.

If point $S$ is situated at infinity then,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mi>W</mi><msup><mi>q</mi><mo>'</mo></msup></mfrac><mo>=</mo><msub><msub><mi>V</mi><mi>r</mi></msub><mn>1</mn></msub><mo>-</mo><mi>V</mi><mfenced><mi>α</mi></mfenced><mspace linebreak="newline"/><mi>o</mi><mi>r</mi><mo>,</mo><mo> </mo><msub><msub><mi>V</mi><mi>r</mi></msub><mn>1</mn></msub><mo>=</mo><mfrac><mi>W</mi><msup><mi>q</mi><mo>'</mo></msup></mfrac></math>
(as Potential energy at infinity is zero)

Therefore, the electric potential at a point in an electric field is the ratio of work done in bringing a test charge from infinity to that point to the magnitude of the test charge.

Potential due to point charge:

We have electric potential at a point due to a test charge $+q$,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>V</mi><mo> </mo><mo>=</mo><mo> </mo><mfrac><mi>q</mi><mrow><mn>4</mn><msub><mi>πε</mi><mn>0</mn></msub><mi>r</mi></mrow></mfrac></math>

Potential $V$ at any point due to all the point charges in a given system is given by,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>V</mi><mo> </mo><mo>=</mo><mo> </mo><mfrac><mn>1</mn><mrow><mn>4</mn><msub><mi>πε</mi><mn>0</mn></msub></mrow></mfrac><munderover><mo>∑</mo><mrow><mi>i</mi><mo>=</mo><mn>1</mn></mrow><mi>n</mi></munderover><mfrac><msub><mi>q</mi><mi>i</mi></msub><msub><mi>r</mi><mi>i</mi></msub></mfrac></math>

For continuous charge distribution, the above equation becomes,

<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>V</mi><mo> </mo><mo>=</mo><mo> </mo><mfrac><mn>1</mn><mrow><mn>4</mn><msub><mi>πε</mi><mn>0</mn></msub></mrow></mfrac><mo>∫</mo><mfrac><mrow><mi>d</mi><mi>q</mi></mrow><mi>r</mi></mfrac></math>

Field and potential gradient:

Let us consider a diagram showing the electric field $E$ due to a point charge $+q$ at point $O in a radially outward direction.

 Relation between electric fiels and electric potential

Figure:7.b

Consider a test charge $q^{‘}$is at point R.

Now force on $q^{‘}$ is,

$F=q^{‘}E$

Work done by the force in displacing $q^{‘}$from $R$ to $S$
is,$dW=F.dr=q^{‘}E.dr$
Change in potential energy is,$dU=-dW=-q^{‘}E.dr$
Change in electric potential is,$dV=dU/q^{‘}$

Or, $dV=-E.dr$

Or, $E=-\frac{dV}{dr}$

The quantity $\frac{dV}{dr}$ is the rate of change of potential with distance. This is known as potential gradient.

Negative sign indicates that the potential drops with increasing distance.

A graph of potential gradient and electric field intensity for a radial field is shown below.

Figure:7.c

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