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Electrostatics question

2015 · Q116
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Electrostatics question

2015 · Q116

NEETPhysicsElectrostaticsMCQ+4 / −1
The electric field in a certain region is acting radially outward and is given by E = Ar. A charge contained in a sphere of radius 'a' centred at the origin of the field, will be given by
  1. A
    4πε0Aa34\pi {\varepsilon _0}A{a^3}4πε0​Aa3
  2. B
    ε0Aa3{\varepsilon _0}A{a^3}ε0​Aa3
  3. C
    4πε0Aa24\pi {\varepsilon _0}A{a^2}4πε0​Aa2
  4. D
    Aε0a2A{\varepsilon _0}{a^2}Aε0​a2
View written solutionFree

Correct answer: A

According to question, electric field varies as

E = Ar

Here r is the radial distance.

At r = a, E = Aa     …(i)

Net flux emitted from a spherical surface of radius a is ϕnet=qenε0{\phi _{net}} = {{{q_{en}}} \over {{\varepsilon _0}}}ϕnet​=ε0​qen​​

⇒(Aa)×(4πa2)=qε0 \Rightarrow \left( {Aa} \right) \times \left( {4\pi {a^2}} \right) = {q \over {{\varepsilon _0}}}⇒(Aa)×(4πa2)=ε0​q​   [Using equation (i)]

∴q=4πε0Aa3 \therefore q = 4\pi {\varepsilon _0}A{a^3}∴q=4πε0​Aa3

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