InterviewSolution
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InterviewSolution
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(a) Using Gauss law, derive expression for electric field due to a spherical shell a uniform charge distribution sigma and radius R at a point lying at a distance r from the centre of shell, such that (i) O lt r lt R, and (ii) r gt R. (b) An electric field is uniform and acts along +x direction in the region of positive x. It is also uniofrm with the same magnitude but acts in -x direction in the region of negative x. The value of the fleld is E= 200 N/C for xgt 0 and E=-200 N//C" for "x lt 0. A right circular cylinder of length 20 cm and radius 5 cm has its centre at the origin and its axis along the x-axis so that one flat face is at x =+ 10 cm and the other is at x=-10 cm. Find : (i) The net outward flux through the cylinder. (ii) The net change present inside the cylinder. |
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Answer» Solution :(a) (i) Consider a hollow charged conducting sphere of radius R and having charge Q. To find electric field at a point P inside the shell, consider a sphere through point P and having centre O, i.e, r= OP `("where "O lt r lt R)` as the GAUSSIAN surface. The electric flux through the Gaussian surface `phi_(E)=oint vecE. hatn ds=E 4pi r^(2) ....(i)` According to Gauss theorem, total electric flux should be `phi_(E)=(1)/(epsi_(0)) ("charged enclosed")=0` Thus, `phi_(E)=E.4pi r^(2)=0` [Since the Gaussian surface is not enclosing any charge) Thus,`phi_(E)=E.4 pi r^(2)=0` `rArr E=0`  (ii) To find electric field intensity at a point Poutside the shell situated at a distance `r (r gt R)` from the centre of shell, consider a sphere of radius r as the Gaussian surface. All points on this surface are equivalent relative to given charged shell and, thus, electric field E at all points of Gaussian surface has same magnitude E and `vecE` and i are parallel to each other. Total electric flux over the Gaussian surface `phi_(E)=oint vecE. hatn dx=E.4pir^(2) ......(i)`  According to Gauss.s theorem, `phi_(E)=1/in_(0) ("charged enclosed")=Q/(in_(0))` ........(ii) Comparing (i) and (ii), we get `E.4pi r^(2)=Q/in_(0) or E=Q/(4pi in_(0) r^(2))` If surface charge DENSITY of shell be `sigma" then Q"=4piR^(2), sigma` and therefore `E=(4piR^(2) sigma)/(4pi in_(0) r^(2))=(sigmaR^(2))/(in r^(2))` (B) The situation is shown in the given figure. Here, `E=200 NC^(-1)` length of cylinder l=20cm, 0.2m, radius of cylinder R=5cm, =0.05m (i) As shown in figure the net OUTWARD electric flux is along x-axis and HASA value `phi_(E)=2E triangles` `=2EpiR^(2)` `=2 xx 200 xx 3.14 xx (0.05)^(2)` `=3.14 Nm^(2) C^(-1)` (ii) As per Gauss law, net charge present inside the cylinder `q=in_(0) xx phi_(E)=(8.854 xx 10^(-12))xx 3.14 =2.78 xx 10^(-11) C` 
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