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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.

Answer

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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 2

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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