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Electrostatics question

2020 · Q153
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Electrostatics question

2020 · Q153

NEETPhysicsElectrostaticsMCQ+4 / −1
A spherical conductor of radius 10 cm has a charge of 3.2 ×\times× 10-7 C distributed uniformly. That is the magnetude of electric field at a point 15 cm from the centre of the sphere?
(14πε0=9×109Nm2/c2)\left( {{1 \over {4\pi {\varepsilon _0}}} = 9 \times {{10}^9}N{m^2}/{c^2}} \right)(4πε0​1​=9×109Nm2/c2)
  1. A
    1.28 ×\times× 105 N/C
  2. B
    1.28 ×\times× 106 N/C
  3. C
    1.28 ×\times× 107 N/C
  4. D
    1.28 ×\times× 104 N/C
View written solutionFree

Correct answer: A

The electric field for the conducting sphere, away from the surface is given by,

E=14πε0Qr2E = {1 \over {4\pi {\varepsilon _0}}}{Q \over {{r^2}}}E=4πε0​1​r2Q​

=9×109×3.2×10−7225×10−4 = {{9 \times {{10}^9} \times 3.2 \times {{10}^{ - 7}}} \over {225 \times {{10}^{ - 4}}}}=225×10−49×109×3.2×10−7​

=0.128×106 = 0.128 \times {10^6}=0.128×106

=1.28×105 = 1.28 \times {10^5}=1.28×105 N/C

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