Deduce an expression for the electric field intensity due to an electric dipole at any point on the space.
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CBSE
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Grade 12
Electric Dipole
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Derive an expression for electric field intensity due to an electric dipole at a point on its axial line.
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Hint: To derive the expression for electric field due to an electric dipole, consider AB is an electric dipole of two point charges – q and + q separated by small distance 2d and then consider that P is a point along the axial line of the dipole at a distance r from the midpoint O of the electric dipole.
Complete step-by-step answer:
Formula used - E=14π∈∘.qx2
We have been asked to derive an expression for electric field intensity due to an electric dipole at a point on its axial line.
Now, let AB be an electric dipole of two-point charges -q and + q separated by small distance 2d.
Also, consider that P is a point along the axial line of the dipole at a distance r from the midpoint O of the electric dipole.
Now, for better understanding refer the figure below-
The electric field at any point due to a charge q at a distance x is given by-
E=14π∈∘.qx2
Now, the electric field at the point P due to + q charge placed at B is given by-
E1=14π∈∘.q(r−d)2 (along BP)
Here, (r – d) is the distance of point P from charge +q.
Also, the electric field at point due to – q charge placed at A-
E2=14π∈∘.q(r+d)2 (along PA)
Therefore, the magnitude of resultant electric field (E) acts in the direction of the vector with a greater magnitude.
The resultant electric field at P is-
E=E1+(−E2)
Putting the values; we get-
E=[14π∈∘.q(r−d)2−14π∈∘.q(r+d)2] (along BP)
Simplifying further we get-
E=q4π∈∘.[1(r−d)2−1(r+d)2]E=q4π∈∘.[(r+d)2−(r−d)2(r−d)2(r+d)2]
Now, we can write (r−d)2(r+d)2=(r2−d2)2
So, we get-
E=q4π∈∘.[(r)2+(d)2+2rd−(r)2−(d)2+2rd(r−d)2(r+d)2]E=q4π∈∘.[4rd(r2−d2)2]
Now, if the point P is far away from the dipole, then d≪r
So, the electric field will be-
E=q4π∈∘.[4rd(r2)2]=q4π∈∘.4rdr4=q4π∈∘.4dr3 along BP
Also, electric dipole moment p=q×2d , so the expression will now become-
E=q4π∈∘.2(2d)r3=q4π∈∘.2pr3
E acts in the direction of dipole moment.
Therefore, the expression for the electric field intensity due to an electric dipole at a point on its axial line is-
E=q4π∈∘.2(2d)r3=q4π∈∘.2pr3
Note: Whenever it is required to derive an expression, then always first draw a rough sketch depicting a dipole and a point on the axial line. As mentioned in the figure, first we found out the electric field at the point P due to + q charge placed at B and then due to charge – q placed at A. Then, the resultant electric field at P is found out. After making necessary assumptions, we got the expression for electric field intensity.
Answer:
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Questions & Answers
CBSE
Physics
Grade 12
Electric Dipole
Question
Answers
Related Questions
Derive an expression for electric field intensity due to an electric dipole at a point on its axial line.
Answer
VerifiedVerified
26.2K+ Views
6 Likes
Hint: To derive the expression for electric field due to an electric dipole, consider AB is an electric dipole of two point charges – q and + q separated by small distance 2d and then consider that P is a point along the axial line of the dipole at a distance r from the midpoint O of the electric dipole.
Complete step-by-step answer:
Formula used - E=14π∈∘.qx2
We have been asked to derive an expression for electric field intensity due to an electric dipole at a point on its axial line.
Now, let AB be an electric dipole of two-point charges -q and + q separated by small distance 2d.
Also, consider that P is a point along the axial line of the dipole at a distance r from the midpoint O of the electric dipole.
Now, for better understanding refer the figure below-
The electric field at any point due to a charge q at a distance x is given by-
E=14π∈∘.qx2
Now, the electric field at the point P due to + q charge placed at B is given by-
E1=14π∈∘.q(r−d)2 (along BP)
Here, (r – d) is the distance of point P from charge +q.
Also, the electric field at point due to – q charge placed at A-
E2=14π∈∘.q(r+d)2 (along PA)
Therefore, the magnitude of resultant electric field (E) acts in the direction of the vector with a greater magnitude.
The resultant electric field at P is-
E=E1+(−E2)
Putting the values; we get-
E=[14π∈∘.q(r−d)2−14π∈∘.q(r+d)2] (along BP)
Simplifying further we get-
E=q4π∈∘.[1(r−d)2−1(r+d)2]E=q4π∈∘.[(r+d)2−(r−d)2(r−d)2(r+d)2]
Now, we can write (r−d)2(r+d)2=(r2−d2)2
So, we get-
E=q4π∈∘.[(r)2+(d)2+2rd−(r)2−(d)2+2rd(r−d)2(r+d)2]E=q4π∈∘.[4rd(r2−d2)2]
Now, if the point P is far away from the dipole, then d≪r
So, the electric field will be-
E=q4π∈∘.[4rd(r2)2]=q4π∈∘.4rdr4=q4π∈∘.4dr3 along BP
Also, electric dipole moment p=q×2d , so the expression will now become-
E=q4π∈∘.2(2d)r3=q4π∈∘.2pr3
E acts in the direction of dipole moment.
Therefore, the expression for the electric field intensity due to an electric dipole at a point on its axial line is-
E=q4π∈∘.2(2d)r3=q4π∈∘.2pr3
Note: Whenever it is required to derive an expression, then always first draw a rough sketch depicting a dipole and a point on the axial line. As mentioned in the figure, first we found out the electric field at the point P due to + q charge placed at B and then due to charge – q placed at A. Then, the resultant electric field at P is found out. After making necessary assumptions, we got the expression for electric field intensity.
26.2K+ Views
6 Likes