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CBSE Class 12 Physics 2014 Outside Delhi Set 1 Paper

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Question : 28 of 30
Marks: +1, -0
Draw a labelled diagram of Van de Graff generator. State its working principle to show how by introducing a small charged sphere into larger sphere, large amount of charge can be transferred to the outer sphere. State the use of this machine and also point out its limitations.
OR
(a) Deduce the expression for the torque acting on a dipole of dipole moment p→\overset{\rightarrow}{p} in the presence of a uniform electric field E⃗\vec{E}.
(b) Consider two hollow concentric spheres, S1S_1 and S2S_2, enclosing charges 2Q2 Q and 4Q4 Q respectively as shown in the figure.
(i) Find out the ratio of the electric flux through them.
(ii) how will the electric flux through the sphere S1S_1 change if a medium of dielectric constant ' εr′\varepsilon_r' ', ' εr\varepsilon_r ', is introduced in the space inside S1S_1 in place of air?
Deduce the necessary expression.
Solution:  
OR
(a) Consider an electric dipole consisting of charges →q\rightarrow q and +q+q and of length 2a2 a placed in a uniform electric field E⃗\vec{E} making an angle θ\theta with electric field.
Force on charge −q-q at A=−qE⃗A=-q \vec{E} (opposite to E⃗\vec{E} )
Force on charge +q+q at B=+qE⃗(B=+q \vec{E}( along E⃗)\vec{E})
Electric dipole is under the action of two equal and unlike parallel force, which give rise to a torque on the dipole.
τ=    Force  ×\tau = \;\;\text{Force}\;\times Perpendicular distance between the two forces
τ=  qE(AN)=qE(2asin⁡θ)\tau = \; q E (A N) = q E (2 a \sin \theta)
τ=  pEsin⁡θ\tau = \; p E \sin \theta       [∵2qa=P]\;\;\;[ \because 2 q a = P ]
τ⃗=  p⃗×E⃗\vec{\tau} = \; \vec{p} \times \vec{E}
(b) (i) Charge enclosed by sphere S1=2QS_1 = 2 Q
By Gauss' law, electric flux through sphere S1S_1 is
ϕ1=  2Qε0\phi_1 = \; \frac{2 Q}{\varepsilon_0}
Charge enclosed by sphere S2S_2 is
Q′=2Q+4Q=6QQ' = 2 Q + 4 Q = 6 Q
Electric flux through sphere S2S_2 is
∴    ϕ2=  6Qε0\therefore \;\; \phi_2 = \; \frac{6 Q}{\varepsilon_0}
The ratio of the electric flux is
  ϕ1ϕ2=  2Qε06Qε0=  26=  13\; \frac{\phi_1}{\phi_2} = \; \frac{\frac{2 Q}{\varepsilon_0}}{\frac{6 Q}{\varepsilon_0}} = \; \frac{2}{6} = \; \frac{1}{3}
(ii) For sphere S1S_1, the electric flux is
∴  ϕ′=  2Qεr\therefore \; \phi' = \; \frac{2 Q}{\varepsilon_r}
⇒  ϕ′ϕ1=  ε0εr\Rightarrow \; \frac{\phi'}{\phi_1} = \; \frac{\varepsilon_0}{\varepsilon_r}
⇒ϕ′=ϕ1×  ε0εr\Rightarrow \phi' = \phi_1 \times \; \frac{\varepsilon_0}{\varepsilon_r}
∵  εr>ε0\because \; \varepsilon_r > \varepsilon_0
∴  ϕ′<ϕ1\therefore \; \phi' < \phi_1
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