question_answer

An electric field is uniform in the positive X-direction for positive x and uniform with the same magnitude in the negative X-direction for negative x. It is given that

\[E=200\,\hat{i}\,N{{C}^{-1}}\] for x > 0

\[E=-\,200\,\hat{i}\,N{{C}^{-1}}\] for x < 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 face is at \[x=+10\text{ }cm\] while other is at \[x=-10\text{ }cm.\]

(i) What is the net outward flux through each flat face?

(ii) What is the net outward flux through the cylinder?

(iii) What is the net charge inside the cylinder?

 

Answer:

            According to the question, a right circular cylinder is shown below: Circular cylinder (i) On the left face \[\Delta S=-\Delta \hat{i}=\pi {{(0.05)}^{2}}\]  \[[\because \Delta S=\pi {{r}^{2}}=\pi {{(0.05)}^{2}}]\] \[\therefore \] The outward flux through left face, \[{{\phi }_{1}}=E\cdot \Delta S=-(200\hat{i})\cdot [-\pi {{(0.05)}^{2}}\hat{i}]\] \[\Rightarrow \]   \[{{\phi }_{1}}=200\times \pi \times {{(0.05)}^{2}}\]     \[[\because \hat{i}\cdot \hat{i}=1]\] \[\Rightarrow \]   \[{{\phi }_{1}}=1.57N\text{-}{{m}^{2}}{{C}^{-1}}\] Similarly outward flux through right face, \[{{\phi }_{2}}=E\cdot \Delta S=(200\hat{i})\cdot [-\pi {{(0.05)}^{2}}\hat{i}]\] \[\Rightarrow \]   \[\phi =1.57N-{{m}^{2}}{{C}^{-1}}\] Flux through each flat surface is 1.57 \[N-{{m}^{2}}{{C}^{-1}}.\] (ii) Again at every point on the curved side of the cylinder \[E\bot \Delta S.\] \[\therefore \] Angle between electric field and surface area is \[90{}^\circ .\] \[\therefore \] Flux through curved surfaces \[{{\phi }_{3}}=E\cdot \Delta S=E\Delta S\cos \,\,90{}^\circ =0\] \[[\because \cos 90{}^\circ =0]\] Net outward flux through the cylinder, \[{{\phi }_{net}}={{\phi }_{1}}+{{\phi }_{2}}+{{\phi }_{3}}=1.57+1.57+0\] \[\Rightarrow \,\,{{\phi }_{net}}=3.14\,\,N\text{-}{{m}^{2}}{{C}^{-1}}\] (iii) By Gauss's theorem, net charge enclosed by cylinder,             \[q={{\varepsilon }_{0}}\,\,{{\phi }_{net}}=8.85\times {{10}^{-12}}\times 3.14\] \[\Rightarrow q=2.78\times {{10}^{-11}}C\]