Q 11 :

The electric potential at the centre of two concentric half rings of radii R1 and R2, having same linear charge density λ is

  • 2λε0

     

  • λ2ε0

     

  • λ4ε0

     

  • λε0

     

(2)

dv=14πε0λ dℓR ;     v∫dv=14πε0λℓR

Potential at centre, V=V2+V1

⇒V=(λ.πR2)4πε0R2+(λ.πR1)4πε0R1=λ2ε0



Q 12 :

Figure shows part of a circuit. What is the potential difference VC-VB= __________  V



(20)

Applying Kirchhoff's Junction Law at E current in wire DE is 8A from D to E. Now further applying Junction Law at D. The current in 3Ω resistance will be 3A towards D.