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.



Q 13 :

Two charged conducting spheres S1 and S2 of radii 8 cm and 18 cm are connected to each other by a wire. After equilibrium is established, the ratio of electric fields on S1 and S2 spheres are ES1 and ES2 respectively. The value of ES1ES2 is _______.               [2026]

  • 32

     

  • 23

     

  • 49

     

  • 94

     

(4)

At equilibrium: V1=V2

kQ18=kQ218    {Vsurface=kQR}

⇒Q1Q2=49

ES1ES2=kQ1r12r22kQ2    {Esurface=kQR2}

ES1ES2=Q1Q2(r2r1)2=49×(94)2=94



Q 14 :

Two metal plates (A, B) are kept horizontally with separation of (12π) cm, with plate A on the top. An atomizer jet sprays oil (density 1.5 g/cm3) droplets of radius 1 mm horizontally. All oil droplets carry a charge 5 nC. The potentials VA and VB are required on plates A and B respectively in order to ensure the droplets do not descend. The values of VA and VB are _______.

(Neglect the air resistance to the droplets and take g=10m/s2)                  [2026]

  • 100 V and 580 V

     

  • 580 V and 100 V

     

  • 60 V and 400 V

     

  • 0 V and −200 V

     

(1)

mg=qE⇒Vρg=qE

43πr3·ρ·g=q·(VB-VA)12π×100

43π10-9×1.510×103×10=5×10-9×ΔV×π×10012

ΔV=48×10=480 V



Q 15 :

The electric potential as a function of x, y is given by V=5(x2-y2) V. The electric field at a point (2, 3) m is _____ V/m.             [2026]

  • (-20i^+30j^)

     

  • (20i^-30j^)

     

  • (20i^+45j^)

     

  • (-4i^+6j^)

     

(1)

V=5(x2-y2) V

V=5x2-5y2,  Ex=-∂V∂x=-10xi^

x=2, y=3,  Ey=-∂V∂y=+10yj^

E→=(-20i^+30j^) N/C