- Note
- Things to remember

The physical quantity that drives electric charge in an electric field is called electric potential. Electric potential at a point inside an electric field is defined as the amount of work done in bringing unit positive charge from infinity to that point (without acceleration).

Mathematically,

$$\text {Electric Potential (V)} = \frac {\text {work done} (w)}{\text {charge moved} (q_o)} $$

$$V = \frac {W}{q_o}$$

It is a scalar quantity having unit JC^{-1 }or volt (V). Its dimensional formula is [ML^{2} T^{-3}A^{-1}].

Suppose charge 'q' is present at point 'O'. B is a point at distance 'r' from the charge 'q' where electric potential 'V' is to be calculated. For this a test charge +q_{o} is placed at point 'A' at distance 'x' from the charge 'q'.

Therefore, experienced by this text charge q_{o} is

$$F = \frac {1}{4 \pi \epsilon _o} \frac {qq_o}{x^2} \dots (i)$$

If the charge 'q_{o}' is displaced by 'dx' from A to B. Then the small work done'dw' against electrostatic force of repulsion is

$$dw= -Fdx$$

$$dw = -\frac {1}{4 \pi \epsilon _o} \frac {qq_o}{x^2}dx \dots (ii)$$

The negative sign indicates force and displacement vectors are opposite.

Now, the total work done 'w' in moving charge 'q_{o}' from infinity to point 'P' is obtained by integrating 'dw' between the limits infinity and 'r'.

$$\therefore w = \int_\infty^r dw$$

$$= \int_\infty^r -\frac {1}{4 \pi \epsilon _o} \frac {qq_o}{x^2}$$

$$w = -\frac {qq_o}{4 \pi \epsilon _o}\int_\infty^r \frac {1}{x^2} dx$$

$$= -\frac {qq_o}{4 \pi \epsilon _o}\int_\infty^r \frac {1}{x^{-2}} dx$$

$$= -\frac {qq_o}{4 \pi \epsilon _o} \left [ \frac {x^{-2+1}}{-2+1} \right ]_\infty^r $$

$$= -\frac {qq_o}{4 \pi \epsilon _o} \left [ \frac {-1}{x}\right ]_\infty^r $$

$$=\frac {qq_o}{4 \pi \epsilon _o} \left (\frac {1}{r} - \frac {1}{\infty} \right )$$

$$ W =\frac {qq_o}{4 \pi \epsilon _o}. \frac {1}{r} $$

Therefore, by the definition of electric potential. Electric Potential at a point \( P(v) = \frac {work done (w)}{charge moved (q_o} \)

$$= \frac { \frac {qq_o}{4 \pi \epsilon _o}. \frac {1}{r}}{q_o}$$

$$ V=\frac {qq_o}{4 \pi \epsilon _o}. \frac {q}{r} $$

This is the electric potential at distance 'r' from the charge 'q'.

The potential difference between two points in an electric field is defined as " The amount of work done in moving unit positive charge from one point to another point." If W_{AB} represents the amount of work done in moving charge 'q_{o}' from A to B denoted by V_{AB} is defined as a potential difference (V_{AB}).

$$\text {Electric Potential} (V_{AB}) = \frac {\text {work done} (W_{AB})}{\text {charge moved} (q_o)} $$

$$\therefore V_AB = \frac {W_{AB}}{q_o}$$

To calculate the potential difference between points A and B. Suppose charge 'q_{o}' is placed at distance 'x' from the charge 'q'. The force experienced by 'q_{o}' is

$$F = \frac {1}{4 \pi \epsilon _o} \frac {qq_o}{x^2} \dots (i)$$

Suppose the charge 'q_{o}' is slightly displaced 'dx', towards 'B' then the small work done 'dw' against electrostatic force of repulsion is

$$dw= -Fdx$$

$$dw = -\frac {1}{4 \pi \epsilon _o} \frac {qq_o}{x^2}dx \dots (ii)$$

Hence, the total work done 'W_{AB}'in moving charge 'q_{o}' from 'A' to 'B' is obtained by integrating 'dw' between the limits 'r'_{A}' and'r_{A}'.

$$\therefore w = \int_{r_A}^{r_B} dw$$

$$= \int_{r_A}^{r_B} -\frac {1}{4 \pi \epsilon _o} \frac {qq_o}{x^2}$$

$$w = -\frac {qq_o}{4 \pi \epsilon _o} \int_{r_A}^{r_B} \frac {1}{x^2} dx$$

$$= -\frac {qq_o}{4 \pi \epsilon _o} \left [ \frac {x^{-2+1}}{-2+1} \right ]_{r_A}^{r_B} $$

$$= -\frac {qq_o}{4 \pi \epsilon _o} \left [ \frac {-1}{x}\right ]_{r_A}^{r_B} $$

$$=\frac {qq_o}{4 \pi \epsilon _o} \left (\frac {1}{r_A} - \frac {1}{r_B} \right )$$

Therefore by the definition of potential difference,

$$\text {Electric Potential} (V_{AB}) = \frac {\text {work done} (W_{AB})}{\text {charge moved} (q_o)} $$

$$= \frac { \frac {qq_o}{4 \pi \epsilon _o}.\left ( \frac {1}{r_B}-\frac {1}{r_A}\right )}{q_o}$$

$$= \frac {qq_o}{4 \pi \epsilon _o}.\left ( \frac {1}{r_B}-\frac {1}{r_A}\right )$$

Above expression indicates the potential difference between two points depends upon the position of the points and the magnitude of the charge producing the field.

**One electron Volt (1 **ev**)**

It is a smaller unit of energy used in atomic and molecular scale.The energy gained by an electron which has been accelerated between two points having a potential difference of 1 volt is known as 1 electron volt.

We know that

$$ \text {energy gained} = \text {work done} $$

$$= q \times V_{AB}$$

When an electron is allowed to move,

$$ q = e = 1.6 \times 10^{-19} C, V_{AB} = 1 \text {volt, the energy gained in one electron volt.} $$

$$ 1 ev = 1.6 \times 10^{-19} \times 1 \text {volt} $$

$$ \therefore 1 ev = 1.6 \times 10^{-19} \text {Joule (J)} $$

Hence, 1 electron volt (1 ev) represents \(1.6\times 10 ^{-19} \) Joule (J).

The physical quantity that drives electric charge in an electric field is called electric potential.

Electric potential at a point inside an electric field is defined as the amount of work done in bringing unit positive charge from infinity to that point (without acceleration).

The potential difference between two points in an electric field is defined as " The amount of work done in moving unit positive charge from one point to another point."

The energy gained by an electron which has been accelerated between two points having a potential difference of 1 volt is known as 1 electron volt.

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## Omprakash Ray Tharu

Why is earth potential is taken as zero?

Feb 08, 2017

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