Answer
Verified
397.5k+ 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, $K=\dfrac{1}{4\pi \epsilon_{0}}$.
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
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Write an application to the principal requesting five class 10 english CBSE
Difference Between Plant Cell and Animal Cell
a Tabulate the differences in the characteristics of class 12 chemistry CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
Discuss what these phrases mean to you A a yellow wood class 9 english CBSE
List some examples of Rabi and Kharif crops class 8 biology CBSE