Correct Answer - Option 4 : E/2

__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

__CALCULATION:__

Given R = radius of the sphere,E_{1} = E, q_{1} = Q, q2 = 2q_{1} = 2Q, r_{1} = r and r_{2} = 2r

Since r > R

So the point is present outside the sphere.

- We know that the electric field intensity at a point outside the metallic sphere is given as,

\(⇒ E=\frac{kq}{r^2}\) -----(1)

Where E = electric field intensity, k = 9 ×10^{9} N-m^{2}/C^{2}, q = charge on the sphere, and r = distance of the point from the center of the sphere

When q1 = Q and r1 = r,

\(⇒ E_{1}=\frac{kq_{1}}{r_{1}^2}\)

\(⇒ E_{1}=\frac{kQ}{r^2}\) -----(2)

When q2 = 2Q and r2 = 2r,

\(⇒ E_{2}=\frac{kq_{2}}{r_{2}^2}\)

\(⇒ E_{2}=\frac{2\times kQ}{(2r)^2}\)

\(⇒ E_{2}=\frac{kQ}{2r^2}\) -----(3)

By equation 2 and equation 3,

\(⇒ E_{2}=\frac{E_{1}}{2}\)

Since E_{1} = E,

\(⇒ E_{2}=\frac{E}{2}\)

- Hence, option 4 is correct.