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Sunday, 11 June 2023

Electric Potential due to Dipole at any Point

Electric Potential due to Dipole at any Point 


 Consider an electric dipole AB. The charges at points A and B are -q and +q respectively and the distance between them is 2a. We have to determine the electric potential at point P from its center O at distance r and the line OP makes angle `\theta` with the axis of the dipole.

Electric Potential due to Dipole at any Point
Electric Potential due to Dipole at any Point


We know that the formula Electric Potential due to a Point Charge is


`V_P = \frac {K  q }{r}`


Electric Potential at P due to charge -q


`V_A = \frac{K (- q)}{r_1}`

 

`V_A = - \frac{K q}{r_1}`


Potential at P due to charge + q

`V_B = \frac{K q}{r_2}`


Thus, the total electric potential at P due to electric dipole


 

`V_P = V_B + V_A`

 

`V_P = \frac{K q}{r_2} + (- frac{K q }{r_1})`

 

`V_P = \frac{K q}{r_2} - frac{K q }{r_1}`

 

`V_P = K q [\frac{1}{r_2} - frac{1 }{r_1}]`

 

`V_P = K q [\frac{r_1 - r_2}{r_1 r_2}]`        ....Eq (1)


Now, to calculate  `(r_1 - r_2)` and `r_1 r_2`.

We draw perpendiculars AC and BD from points A and B online OP respectively.


     If r >>> a, then AP   `\approx`  PC and PB `\approx`  PD

 

Here,

 

OP = OD + DP


`r = a cos\theta + r_2`

 

`r_2 = r - a cos\theta`


Again,


`AP = OP + OC`


`r_1 = r + a cos\theta `


Now, substituting values of `r_1` and `r_2` in eq. (1)


`V_P = K q [\frac{r_1 - r_2}{r_1 r_2}]`


`V_P = K q [\frac{(r + a cos\theta) - (r - a cos\theta)}{(r + a cos\theta)(r - a cos\theta)}]`


`V_P = K q [\frac{r + a cos\theta - r + a cos\theta}{(r - a cos\theta)^2}]`


`V_P = K q [\frac{ a cos\theta + a cos\theta}{(r - a cos\theta)^2}]`


`V_P = K q [\frac{ 2 a cos\theta }{r^2 - a^2 cos^2\theta}]`


`V_P = \frac{K q (2 a) cos\theta }{r^2 - a^2 cos^2\theta}`


`V_P = \frac{K p cos\theta }{r^2 - a^2 cos^2\theta}`    `(\because p = q \times 2a)`


`V_P = \frac{K p cos\theta }{r^2}`    `(\because r^2`>>>`a^2)`


     It is clear that potential due to dipole depends on `r^{-2}` and not on `r^{-1}`.


In the vector form


`V = \frac{K \vecP . \hat r}{r^2}`


`V = \frac{K \vecP . \vec r}{r^3}`        `(\because \hat r = \frac{\vec r}{r})`


Special Cases


Case I -


    If point P is on the axis when `\theta = 0^\circ`, `cos\theta = 1`, then


`V = \frac {K P (1)}{r^2}`


`V = \frac {K P}{r^2}`


Case II -


    If point P is on the equatorial line when `\theta = 90^\circ`, `cos\theta = 0`, then


`V = \frac {K P (0)}{r^2}`


`V = 0`


    Thus, the electric potential is zero due to a dipole on equatorial line.


Important Points


1.    For a small electric dipole, the electric potential is inversely proportional to the square of the distance.


2.    The potential due to an electric dipole does not depend on the distance only but it also depends upon the angle between the displacement vector `\vecr` and the dipole moment.


3.    The electric potential is zero due to a dipole on the equatorial line.


Example


    Two point charges `8 \times 10^{-19} C` and `- 8 \times 10^{-19} C` are at `2 \times 10^{-10} m`. Determine the electric potential at the distance `4 \times 10^{-6} m` from this dipole when the point


(a)    is at the axis of dipole


(b)    is at equatorial line of dipole


(c)    is at `60^\circ` from the dipole moment.


Here,

`q = 8 \times 10^{-19} C`

`2 a = 2 \times 10^{-10} m`

` r = 4 \times 10^{- 6} m`

Thus, dipole moment, 

`p = q(2a)`

`p = 8 \times 10^{-19} \times 2 \times 10^{-10}`

`p = 16 \times 10^{-29} C.m`

(a)    Electric potential at axis


`V = \frac {K P}{r^2}`


`V = \frac {9 \times 10^9 \times 16 \times 10^{- 29}}{(4 \times 10^{-6})^2}`


`V = 9 \times 10^{-8} ` volt


(b)    At equatorial line of dipole 


`\theta = 90^\circ`, `cos\theta = 0`, then


`V = \frac {K P (0)}{r^2}`


`V = 0`


(c)    If `\theta = 60^\circ` then `cos 60^\circ = \frac{1}{2}`

`V_P = \frac{K p cos60^\circ }{r^2}` 

`V = \frac {9 \times 10^9 \times 16 \times 10^{- 29} \frac{1}{2}}{(4 \times 10^{-6})^2}`


`V = 4.5 \times 10^{-8} ` volt



Short Answer Type Questions with Answers


1. What is an electric dipole?

Answer: An electric dipole consists of two equal and opposite charges separated by a very small distance like `NaCl`.


2. What is the formula for electric dipole moment?

Answer: The formula for electric dipole moment is given by p = q(2a), where q is the magnitude of the charge and 2a is the separation between the charges.


3. How is the electric field due to an electric dipole defined?

Answer: The electric field due to an electric dipole is defined as the force experienced by a positive test charge placed at a point in the electric field.


4. What is the expression for the electric potential due to an electric dipole at a point on its axial line?

Answer: The expression for the electric potential due to an electric dipole at any point is `V = \frac{K p cos\theta}{r^2}`, where K is the electrostatic constant `(K = 9 \times 10^9)\frac{N m^2}{C^2}`, p is the electric dipole moment, θ is the angle between the dipole moment and the line connecting the point to the dipole, and r is the distance between the point and the dipole.


5. How does the electric potential due to an electric dipole vairy on its equatorial line?

Answer: On the equatorial line of an electric dipole, the electric potential is zero.


6. What is the relation between the electric potential and electric field due to an electric dipole?

Answer: The electric potential due to an electric dipole is related to the electric field by the equation `dV = - \vecE.d\vecr`, where V is the electric potential, E is the electric field, and r is the distance from the dipole.


7. Can the electric potential due to an electric dipole be negative?

Answer: Yes, the electric potential due to an electric dipole can be negative. It depends on the orientation of the dipole and the position of the point at which the potential is measured.


8. How does the electric potential due to an electric dipole vairy with distance?

Answer: The electric potential due to an electric dipole decreases as the distance from the dipole increases `(V \prop r^{-2})`.


9. What happens to the electric potential due to an electric dipole at large distances?

Answer: At large distances, the electric potential due to an electric dipole approaches zero.


10. What is the significance of the electric potential due to an electric dipole?

Answer: The electric potential due to an electric dipole helps in understanding the behavior of charges and their interactions in electric fields. It is used to analyze and calculate the potential energy and forces experienced by charges in the presence of a dipole.





Chapter  2:  ELECTROSTATIC POTENTIAL AND CAPACITANCE  

 

PHYSICS NOTES


























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