JEE Main2013PhysicsElectrostaticsActual
A point charge of magnitude +1 C is fixed at (0 , 0,0) . An isolated uncharged spherical conductor, is fixed with its center at (4,0,0) . The potential and the induced electric field at the centre of the sphere is :
Options
- A1.8 10^5 ~V and -5.625 10^6 ~V / m
- B0 ~V and 0 ~V / m
- C2.25 10^5 ~V and -5.625 10^6 ~V / m
- D2.25 10^5 ~V and 0 ~V / m
Correct answer
C. 2.25 10^5 ~V and -5.625 10^6 ~V / m
Step-by-step solution
q =1 C =1 10⁻⁶ C r =4 ~cm =4 10⁻² ~m Potential V = kq r aligned & = 9 10^9 10⁻⁶ 4 10⁻² & =2.25 10^5 ~V . aligned Induced electric field E =- kq r ^2 = 9 10^9 1 10⁻⁶ 16 10⁻⁴ =-5.625 10^6 ~V / m