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Electrostatics question

2008 · Q166
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Electrostatics question

2008 · Q166

NEETPhysicsElectrostaticsMCQ+4 / −1
The electric potential at a point in free space due to charge Q coulomb is Q ×\times× 1011 volts. The electric field at that point is
  1. A
    4πε0Q×10204\pi {\varepsilon _0}Q \times {10^{20}}4πε0​Q×1020 volt/m
  2. B
    12π\piπ0Q ×\times× 1022 volt/m
  3. C
    4πε0Q×10224\pi {\varepsilon _0}Q \times {10^{22}}4πε0​Q×1022 volt/m
  4. D
    12πε0Q×102012\pi {\varepsilon _0}Q \times {10^{20}}12πε0​Q×1020 volt/m
View written solutionFree

Correct answer: C

V=14πε0×QRV = {1 \over {4\pi {\varepsilon _0}}} \times {Q \over R}V=4πε0​1​×RQ​

= Q × 1011 volt    …(i)

E=14πε0×QR2E = {1 \over {4\pi {\varepsilon _0}}} \times {Q \over {{R^2}}}E=4πε0​1​×R2Q​

= V/R = Q × 1011 × 4πε0{4\pi {\varepsilon _0}}4πε0​ × 1011    from ...(i)

= 4πε0{4\pi {\varepsilon _0}}4πε0​ × Q × 1022 volt/m

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