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The charge on the metallic sphere of radius R is Q. The electric field intensity at a distance r from the centre of the sphere is E. If the charge on the sphere is doubled, the electric field intensity at a distance 2r will be: (R < r)
1. 2E
2. E
3. E/4
4. E/2

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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,E1 = E, q1 = Q, q2 = 2q1 = 2Q, r1 = r and r2 = 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 ×109 N-m2/C2, 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 E1 = E,

\(⇒ E_{2}=\frac{E}{2}\)

  • Hence, option 4 is correct.

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