Derive an expression for the electric field $E$ due to a dipole of length $'2a'$ at a point distant $r$ from the center of the dipole on the axial line.
Answer
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Hint Here, in this question, we have an electric field given, and also the dipole length is given. We have to find the expression for the electric field $E$ on the axial line of an electric dipole. So here we will use the idea of the electric field, and we will be able to get it.
Complete step by step solution:
Firstly we will see the electric field which is on the dipole having an axial line.
So we assume $P$ , at a distance $r$ and the charge $q$
Then, the electric field at the charge which is $ - q$ will be
$ \Rightarrow {E_{ - q}} = \dfrac{q}{{4\pi {\varepsilon _0}{{\left( {r + a} \right)}^2}}}\mathop p\limits^ \wedge $
Here,
$\mathop p\limits^ \wedge $ , The unit vector along the dipole axis $\left( {from \,{{ - q \,to \, + q}}} \right)$ .
Now, we will see the electric field at $ + q$ will be
$ \Rightarrow {E_{ + q}} = \dfrac{q}{{4\pi {\varepsilon _0}{{\left( {r - a} \right)}^2}}}\mathop p\limits^ \wedge $
Therefore, now we will find the total electric field that is the sum of both the above electric fields.
$ \Rightarrow E = {E_{ + q}} + {E_{ - q}}$
Now on substituting the values we get from both the electric field
$ \Rightarrow \dfrac{q}{{4\pi {\varepsilon _0}}}\left[ {\dfrac{1}{{{{\left( {r - a} \right)}^2}}} - \dfrac{1}{{{{\left( {r + a} \right)}^2}}}} \right]\mathop p\limits^ \wedge $
Now on again solving the above equations, we get
$ \Rightarrow E = \dfrac{q}{{4\pi {\varepsilon _0}}}\dfrac{{4ar}}{{{{\left( {{r^2} - {a^2}} \right)}^2}}}\mathop p\limits^ \wedge $
And if $r > > a$ then,
$ \Rightarrow \dfrac{{4qa}}{{4\pi {\varepsilon _0}{r^3}}}\mathop p\limits^ \wedge {{ }}\left( {r > > a} \right)$
And it can also be written as
$ \Rightarrow E = \dfrac{{2p}}{{4\pi {\varepsilon _0}{r^3}}}{{ }}\left( {\sin {{ce \vec p = q}} \times {{ 2\vec a}}\mathop p\limits^ \wedge } \right)$
Therefore, $\dfrac{q}{{4\pi {\varepsilon _0}}}\dfrac{{4ar}}{{{{\left( {{r^2} - {a^2}} \right)}^2}}}\mathop p\limits^ \wedge $ will be the expression for an electric field.
Note: If the electric field is constant, this would indicate that neither the direction nor the magnitude is changing concerning time or space. Any electric charge experiences a force when it is in an electric field. Hence, we say that a region of space has an electric field if a charge experiences a force in this region. I believe an electric field exists within our perceptions. That is how, by creating another point in space that will serve as a reference point that causes its existence. A point in space that could be similar to any other air molecules being or becoming electromagnetically or statically charged possessing a magnitude and directions.
Complete step by step solution:
Firstly we will see the electric field which is on the dipole having an axial line.
So we assume $P$ , at a distance $r$ and the charge $q$
Then, the electric field at the charge which is $ - q$ will be
$ \Rightarrow {E_{ - q}} = \dfrac{q}{{4\pi {\varepsilon _0}{{\left( {r + a} \right)}^2}}}\mathop p\limits^ \wedge $
Here,
$\mathop p\limits^ \wedge $ , The unit vector along the dipole axis $\left( {from \,{{ - q \,to \, + q}}} \right)$ .
Now, we will see the electric field at $ + q$ will be
$ \Rightarrow {E_{ + q}} = \dfrac{q}{{4\pi {\varepsilon _0}{{\left( {r - a} \right)}^2}}}\mathop p\limits^ \wedge $
Therefore, now we will find the total electric field that is the sum of both the above electric fields.
$ \Rightarrow E = {E_{ + q}} + {E_{ - q}}$
Now on substituting the values we get from both the electric field
$ \Rightarrow \dfrac{q}{{4\pi {\varepsilon _0}}}\left[ {\dfrac{1}{{{{\left( {r - a} \right)}^2}}} - \dfrac{1}{{{{\left( {r + a} \right)}^2}}}} \right]\mathop p\limits^ \wedge $
Now on again solving the above equations, we get
$ \Rightarrow E = \dfrac{q}{{4\pi {\varepsilon _0}}}\dfrac{{4ar}}{{{{\left( {{r^2} - {a^2}} \right)}^2}}}\mathop p\limits^ \wedge $
And if $r > > a$ then,
$ \Rightarrow \dfrac{{4qa}}{{4\pi {\varepsilon _0}{r^3}}}\mathop p\limits^ \wedge {{ }}\left( {r > > a} \right)$
And it can also be written as
$ \Rightarrow E = \dfrac{{2p}}{{4\pi {\varepsilon _0}{r^3}}}{{ }}\left( {\sin {{ce \vec p = q}} \times {{ 2\vec a}}\mathop p\limits^ \wedge } \right)$
Therefore, $\dfrac{q}{{4\pi {\varepsilon _0}}}\dfrac{{4ar}}{{{{\left( {{r^2} - {a^2}} \right)}^2}}}\mathop p\limits^ \wedge $ will be the expression for an electric field.
Note: If the electric field is constant, this would indicate that neither the direction nor the magnitude is changing concerning time or space. Any electric charge experiences a force when it is in an electric field. Hence, we say that a region of space has an electric field if a charge experiences a force in this region. I believe an electric field exists within our perceptions. That is how, by creating another point in space that will serve as a reference point that causes its existence. A point in space that could be similar to any other air molecules being or becoming electromagnetically or statically charged possessing a magnitude and directions.
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