The correct answer is : 3
At an equatorial point of a dipole, the electric field due to the dipole is given by:
\(E = k [\frac{p}{ (r^3)}]\)
where k is the Coulomb constant, p is the dipole moment, and r is the distance from the dipole.
For an equatorial point, the distance r is equal to the distance between the two charges of the dipole, which is 2a, where a is the separation between the charges.
Therefore, the electric field at the equatorial point can be written as:
\(E = k [\frac{p}{ (2a)^3}]\)
\(E = (\frac{1}{4πε0}) [\frac{2psinθ}{ (2a)^3}]\)
where θ is the angle between the axis of the dipole and the line joining the dipole to the point where the electric field is to be calculated, and ε0 is the permittivity of free space.
Since sinθ = 1 at the equatorial point, we can simplify the above equation as:
\(E = (\frac{1}{4πε0)} [\frac{psinθ}{ a^3}]\)
\(E = (\frac{1}{4πε0}) [\frac{p}{ a^3}]\)
Therefore, the dependence of the electric field on the distance r at an equatorial point is \(\frac{1}{r^3}\), which is equivalent to n = 3.
Hence, the value of n is 3.
The correct answer is : 3
\(\vec{E}=\frac{KP}{r^3}\)
\(⇒E∝\frac{1}{r^3}\)
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