JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Electrostatics

The volume charge density of a sphere of radius 6 m is

2 μ C cm^{- 3}

. The number of lines of force per unit surface area coming out from the surface of the sphere is

____

\times 1 0^{10} NC^{- 1}

.
[Given : Permittivity of vacuum

ϵ_{0} = 8.85 \times 1 0^{- 12} C^{2} N^{- 1} - m^{- 2}

]

Answer: 45

Solution

The number of electric field lines per unit area equals the electric field magnitude at the surface.

For a uniformly charged sphere with volume charge density

ρ

, the electric field at the surface (radius

R

) is:

E = \frac{ρR}{3 ϵ _{0}}

Converting

ρ = 2 μ C/cm^{3} = 2 \times 1 0^{- 6} \times 1 0^{6} = 2 C/m^{3}

and

R = 6

E = \frac{2 \times 6}{3 \times 8.85 \times 1 0 ^{- 12}} = \frac{12}{26.55 \times 1 0 ^{- 12}} = 0.45 \times 1 0^{12} = 45 \times 1 0^{10} N/C

Therefore, the answer is 45.

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Integer answer

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JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Electrostatics

Solution

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