Question

what. is the ratio of strength of electric field at a point on axial line and a point on same distance a equitorial line of an electric dipole of very small length​

Answer 1

We know that the axial line of a dipole is the line passing through the positive and negative charges of the electric dipole.

Therefore,

The electric field at axial line =

$E_{\text{axial}}=\frac{1}{4\times\pi\times\varepsilon_{\circ}}\times\frac{2P}{r^{3}}$

The electric field at equatorial line =

$E_{\text{equatorial}}=\frac{1}{4\times\pi\times\varepsilon_{\circ}}\times\frac{P}{r^{3}}$

Therefore, the ratio of electric field intensity at a point on the equatorial line to the field at a point on axial line

$\frac{E_{\text{axial}}}{E_{\text{equatorial}}}=\frac{\frac{1}{4\times\pi\times\varepsilon_{\circ}}\times\frac{2P}{r^{3}}}{\frac{1}{4\times\pi\times\varepsilon_{\circ}}\times\frac{P}{r^{3}}}= \frac{2}{1} \\ \\ { E_{axial} } \: E_{equatorial} = 2 \: 1$

Answer 2

The ratio of the strength of an electric field at a point on axial line and a point on same distance at an equitorial line of an electric dipole of very small length is 2:1

- A force which would be exerted on other charged particles in a region around a charged particle object is called electric field

- The formula is given by equation,

E = 1 × 2P /4 πε0 r^3

- With the help of axial line of dipole the other two charges , i.e , positive and negative charges pass through it.

- The electric field at the axial line is equal to,

E(axial) = 1 × 2P /4 πε0 r^3

- The electric field at equatorial line is equal to,

E(equatorial)= 1 × P /4 πε0 r^3

- The ratio of the electric field intensity at the point on the equatorial line to the electric field at the point on axial line is,

- E(axial)/E(equatorial) =

1 × 2P /4 πε0 r^3/ 1 × P /4 πε0r^3

= 2/1

- *E(axial) : E(equatorial) = 2:1*