Question

drive and expression for field along
the axial line of a dipole​

Answer 1

Explanation:

✨✨Explain axial field or electric field on the axial line of an electric dipole i.e. end on position✨✨

A line passing through the positive and negative charges of the dipole is called the axial line of the electric dipole. Consider a point P on the axial line of a dipole (situated in a vacuum) of length 2a at a distance r from the midpoint O.

Electric intensity at P due to –q charge at A, i.e.,

E1=kqAP2=kq(r+a)2

It is represented in magnitude and direction by PQ−→−, i.e., E1.−→ Electric intensity at P due to +q charge at B, i.e.,

E2=kqBP2=kq(r−a)2

It is represented in magnitude and direction by PR−→−, i.e., E2.−→ Let E→ be the resultant electric intensity at P.

According to the principle of superposition of electric fields, E→=E1−→+E2−→

Since E1−→ and E2 act in opposite directions, the resultant electric intensity (E→) at P due to the dipole is represented by PS−→. Clearly, as ∣∣∣E2−→∣∣∣>∣∣∣E1−→∣∣∣, E=E2−E1=k[q(r−a)2−q(r+a)2] =kq[(r+a)2−(r−a)2(r−a)2(r+a)2]=kq4ra(r2−a2)2

Or E=k2×2qa×r(r2−a2)2=k2pr(r2−a2)2

Since E→andp→ are in the same direction,

E→=k2p→r(r2−a2)2=14π∈02p→r(r2−a2)2 … (1)

Special cases:

If r is very large as compared to a, i.e., if the dipole is short, a2 can be neglected as compared to r2 and as such,

E→=k2p→r3=14π∈02p→r3 … (2)

E varies as 1/r3 in case of an electric dipole whereas in case of monopole, E∝1r.

If the electric dipole is situated in a medium of relative permittivity ∈r, then

E→=k2p→r∈r(r2−a2)2=14π∈0∈r2p→r(r2−a2)2

and E→=k2p→r∈rr3=14π∈0∈r2p→r3 (Short dipole) …

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Answer 2

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