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Electrostatics question

2021 · 27 Aug · Shift 1 · Q41
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  5. /2021 · 27 Aug · Shift 1 · Q41

Electrostatics question

2021 · 27 Aug · Shift 1 · Q41

JEE MainPhysicsElectrostaticsMCQ+4 / −1
A uniformly charged disc of radius R having surface charge density σ\sigmaσ is placed in the xy plane with its center at the origin. Find the electric field intensity along the z-axis at a distance Z from origin :-
  1. A
    E=σ2ε0(1−Z(Z2+R2)1/2)E = {\sigma \over {2{\varepsilon _0}}}\left( {1 - {Z \over {{{({Z^2} + {R^2})}^{1/2}}}}} \right)E=2ε0​σ​(1−(Z2+R2)1/2Z​)
  2. B
    E=σ2ε0(1+Z(Z2+R2)1/2)E = {\sigma \over {2{\varepsilon _0}}}\left( {1 + {Z \over {{{({Z^2} + {R^2})}^{1/2}}}}} \right)E=2ε0​σ​(1+(Z2+R2)1/2Z​)
  3. C
    E=2ε0σ(1(Z2+R2)1/2+Z)E = {{2{\varepsilon _0}} \over \sigma }\left( {{1 \over {{{({Z^2} + {R^2})}^{1/2}}}} + Z} \right)E=σ2ε0​​((Z2+R2)1/21​+Z)
  4. D
    E=σ2ε0(1(Z2+R2)+1Z2)E = {\sigma \over {2{\varepsilon _0}}}\left( {{1 \over {({Z^2} + {R^2})}} + {1 \over {{Z^2}}}} \right)E=2ε0​σ​((Z2+R2)1​+Z21​)
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