JEE Main 2007Annual, Paper 1PhysicsMediumElectrostatics167× since 2002The 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 byAOption A. (10/9)(10/9)(10/9) volt / μ\muμ mmm and in the +ve+ ve+ve xxx directionBOption B. (5/3)\left( {5/3} \right)(5/3) volt/ μ\muμ mmm and in the −ve-ve−ve xxx directionCOption C. (5/3)\left( {5/3} \right)(5/3) volt/μ\muμ mmm and in the +ve+ve+ve xxx directionDOption D. (10/9)\left( {10/9} \right)(10/9) volt/ μ m\mu \,mμm and in the −ve-ve−ve xxx directionCheckSkip · show solution
