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Electrostatics question

2007 · Shift 0 · Q82
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Electrostatics question

2007 · Shift 0 · Q82

JEE MainPhysicsElectrostaticsMCQ+4 / −1
The potential at a point xxx(measured in μ m\mu \,mμm) due to some charges situated on the xxx-axis is given by V(x)=20/(x2−4)V\left( x \right) = 20/\left( {{x^2} - 4} \right)V(x)=20/(x2−4) volt The electric field EEE at x=4 μ mx = 4\,\mu \,mx=4μm is given by
  1. A
    (10/9)(10/9)(10/9) volt /μm\mu mμm and in the +vex+ vex+vex direction
  2. B
    (5/3)\left( {5/3} \right)(5/3) volt/μm\mu mμm and in the −vex-vex−vex direction
  3. C
    (5/3)\left( {5/3} \right)(5/3) volt/μm\mu mμm and in the +vex+vex+vex direction
  4. D
    (10/9)\left( {10/9} \right)(10/9) volt/μ m\mu \,mμm and in the −vex-vex−vex direction
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