AP EAMCET20225 Jul 2022Evening ShiftPhysicsElectrostaticsActual
In a space having electric field E =A(x i +y j ) the potential at a point (10 ~m , 20 ~m ) is zero, then the potential at the origin is [ A =10 Vm ⁻² ]
Options
- A500 ~V
- B2000 ~V
- C2500 ~V
- D1500 ~V
Correct answer
C. 2500 ~V
Step-by-step solution
We have, aligned & E =A(x i +y j ) & V=- E d r=-A x d x-A y d y aligned aligned & =- A x ^2 2 - A y ^2 2 + V ₀ & ~V ₀= Integration constant aligned =- A 2 (x^2+y^2 )+V₀=-5 (x^2+y^2 )+V₀ V| _ (10,20) =0 -5 (10^2+20^2 )+ V ₀=0 V ₀=2500 So, . V |_ x =0, y =0 =-5 (0^2+0^2 )+2500=2500 ~V