May 2011 - Career Point

Hence, E x = –∴∂V∂ xE = – ayi – axj= – a[yi + xj]∂V= –ay, E y = –∂ y= – ax3. A charge Q is distributed over two concentric hollowspheres of radii r and R (> r) such the surfacedensities are equal. Find the potential at the commoncentre.q′ qORSol. Let q and q′ be the charges on inner and outer sphere.Thenq + q′ = Q…(1)As the surface densities are equal, henceq q'=2 24πr4πR(∴ Surface density = charge/area)∴ q R 2 = q′ r 2…(2)From eq. (1) q′ = (Q – q), henceq R 2 = (Q – q)r 2q(R 2 + r 2 ) = Q r 222Q rQ R∴ q = and q′ = Q – q =2 22 2R + rR + rNow potential at O is given by1 q 1 q'V = +4πε0r 4πε0r==14πε002Q r 1+2 2(R + r ) r 4πεQ (r + R)4πε (R2 2+ r )r0(RQ r222+ r ) r4. S 1 and S 2 are two parallel concentric spheres enclosingcharges q and 2q respectively as shown in fig.(a) What is the ratio of electric flux through S 1 and S 2 ?(b) How will the electric flux through the sphere S 1change, if a medium of dielectric constant 5 isintroduced in the space inside S 1 in place of air ?S 2Sol. (a) Let Φ 1 and Φ 2 be the electric flux through spheresS 1 and S 2 respectively.q2qS 1∴qΦ 1 =ε0q / εΦ 1 =0Φ 2 3q / ε0and Φ 2 == 31q + 2q 3q=ε ε(b) Let E be the electric field intensity on the surface ofsphere S 1 due to charge q placed inside the sphere.When dielectric medium of dielectric constant K isintroduced inside sphere S 1 , then electric fieldintensity E′ is given byE′ = E/KNow the flux Φ′ through S 1 becomes' 1 qΦ′ =∫E .dS =∫E.dS =K Kε∴ Φ′ =q5ε05. A charge of 4 × 10 –8 C is distributed uniformly on thesurface of a sphere of radius 1 cm. It is covered by aconcentric, hollow conducting sphere of a radius5 cm. (a) Find the electric field at a point 2 cm awayfrom the centre. (b) A charge of 6 × 10 –8 C is placedon the hollow sphere. Find the surface charge densityon the outer surface of the hollow sphere.Sol. (a) See fig. (a) Let P be a point where we have tocalculate the electric field. We draw a Gaussiansurface (shown dotted) through point P. The fluxthrough this surface isq = 6 × 10 –8 C5cmΦ =2cmPFig. (a) Fig. (b)∫∫E.dS = E dS = 4π(2×10 ) E000−22According to Gauss's law, Φ = q/ε 0∴ 4π × (2 × 10 –2 ) 2 E = q/ε 09−8q (9×10 ) × (4×10 )or E ==−22−44πε0 × (2×10 ) 4×10= 9 × 10 5 N/C(b) See fig. (b) We draw a Gaussian surface (showndotted) through the material of hollow sphere. Weknow that the electric field in a conducting material iszero, therefore the flux through this Gaussian surfaceis zero. Using Gauss's law, the total charge enclosedmust be zero. So, the charge on the inner surface ofhollow sphere is 6 × 10 –8 C. So, the charge on theouter surface will be 10 × 10 –8 C.XtraEdge for IIT-JEE 24 MAY 2011