If the ratio of the distance of the two points A and B from a uniformly charged long straight wire is 1 : 2, then the ratio of the electric field intensity at the point A to point B is:

CONCEPT:

Electric field intensity: 

• It is defined as the force experienced by a unit positive test charge in the electric field at any point.

\(⇒ E=\frac{F}{q_{o}}\)    

Where E = electric field intensity, qo = charge on the particle

Electric field intensity due to an infinitely long straight uniformly charged wire:

• The direction of the electric field at every point must be radial (outward if λ > 0, inward if λ < 0).

\(\Rightarrow E=\frac{λ}{2\piϵ_or}\)

Where λ = linear charge density, ϵo = permittivity, and r = distance of the point from the wire

CALCULATION:

Given \(\frac{r_A}{r_B}=\frac{1}{2}\)

• We know that the electric field intensity due to an infinitely long straight uniformly charged wire is given as,

\(\Rightarrow E=\frac{λ}{2\piϵ_or}\)     -----(1)

Where λ = linear charge density, ϵo = permittivity, and r = distance of the point from the wire

By equation 1 the electric field intensity at point A is given as,

\(\Rightarrow E_A=\frac{λ}{2\piϵ_or_A}\)     -----(2)

By equation 1 the electric field intensity at point B is given as,

\(\Rightarrow E_B=\frac{λ}{2\piϵ_or_B}\)     -----(3)

By equation 2 and equation 3,

\(\Rightarrow \frac{E_A}{E_B}=\frac{λ}{2\piϵ_or_A}\times\frac{2\piϵ_or_B}{λ}\)

\(\Rightarrow \frac{E_A}{E_B}=\frac{r_B}{r_A}\)

\(\Rightarrow \frac{E_A}{E_B}=\frac{2}{1}\)

• Hence, option 2 is correct.