# Derive an expression for the electric potential at a point along the axial line of an electric dipole. At a point charge, the values of the electric field and potential are $25N/C$and $10J/C$. Then calculate:

A. the magnitude of the charge

B. distance of the charge from the point of observation

Answer

Verified

206.4k+ views

**Hint:**We know that the electric potential due to a charge, is defined as the amount of energy needed to move a unit positive charge to infinity. Also the potential at any point is the vector sum of potentials at that point. Also, potential is proportional to the charge and inversely proportional to the square of the distance between the point and the charge.

**Formula used:**

$x=\dfrac{V_{X}}{E_{X}}$ and $V=\dfrac{Kq}{r}$

**Complete step by step answer:**

Consider an electric dipole \[AB\] of length $a$ and let $X$ be a point on the axial line such that it is at a distance $x$ from the centre of the dipole \[AB\].

Then the electric potential at the point $X$ is given as,$V_{X}=V_{+q}+V_{-q}$

Or, $V_{X}=K\left[\dfrac{q}{x-a}+\dfrac{-q}{x+a}\right]=\dfrac{K(2qa)}{x^{2}-a^{2}}=\dfrac{KP}{x^{2}-a^{2}}$, where $P$ is the dipole moment of the electric dipole \[AB\]of length $a$ and $K$ is the coulombs constant which is equal to $9\times 10^{9}Nm^{2}/C$

Given that , the electric potential $V_{X}=10J/C$ and the electric field $E_{X}= 25N/C$

Then we know that, $E=\dfrac{V}{r}$ where $r$ is the distance of the charge, substituting the values, we get, $x=\dfrac{V_{X}}{E_{X}}$

Then, we get $x=\dfrac{10}{25}=0.4m$

i.e the distance of the $X$ from the from the centre is \[0.4m\]

Also from the definition of electric potential, we know that , $V=\dfrac{Kq}{r}$ , then substituting the values we get,

$q=\dfrac{V_{X}x}{K}=\dfrac{10\times 0.4}{9\times 10^{9}}=4.45 \times 10^{-10}C$

i.e. the magnitude of each charge is $4.45\times 10^{-10}C$

Hence the answers are as follows:

A. the magnitude of each charge is $4.45\times 10^{-10}C$ and

B. the distance of the $X$ from the from the centre is \[0.4m\]

**Note:**

To find the charge you can also use the $E_{X}$ as $E_{X}=\dfrac{Kq}{x^{2}}$. Also, the value of charge $q$ calculated here, is only one charge of the dipole. Note the dipole moment $P=2aq$. This is taken for simplification and easy calculations. Also, $K=\dfrac{1}{4\pi \epsilon_{0}}$.

Recently Updated Pages

What does the term LOS communication mean Name the class 12 physics CBSE

How do electromagnetic waves travel in a vacuum class 12 physics CBSE

How are gas particles described according to the kinetic class 12 physics CBSE

A ball bounces to 80 of its original height What fraction class 12 physics CBSE

A concave mirror is kept as shown in figure Its principal class 12 physics CBSE

A parallel beam of light enters a clear plastic bead class 12 physics CBSE

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE