question_answer

(i) A 2m insulating slab with a large aluminium sheet of area \[/{{m}^{2}}\]  on its top is fixed by a man outside his house one evening. Will he get an   electric shock, if he touches the metal sheet next morning?

(ii) Two identical capacitors of plate dimensions \[1\times b\] and plate separation d have dielectric slabs filled in between the space of the plates as shown in the figures.

Find the ratio capacitance in each case. Or State Gauss's law in electrostatics. The electric field components in figure shown are \[{{E}_{x}}=\alpha {{x}^{1/2}},\]  \[{{E}_{y}}={{E}_{2}}=0\] in which \[\alpha =800N/C{{m}^{1/2}}.\]

Calulate

(i) the electric flux through the cube

(ii) the charge within the cube

Where the side of cube = 0.1 m

Answer:

Or (i) \[1.05N-{{m}^{2}}{{C}^{-1}}\] (ii) \[9.2\times {{10}^{12}}C\] (i) Yes, the man will get an electric shock when he touches the metal sheet. This is because the steady discharging current in the atmosphere charges up the aluminum sheet. The rise in voltage in the sheet depends on the value of capacitance formed by aluminum sheet, the slab and ground. (ii) In the first case, capacitance is \[{{C}_{1}}=\frac{k{{\varepsilon }_{0}}(l\times b)}{d}\] In the second case, it forms a parallel combination of 2 capacitors. Each with area \[\frac{l}{2}\times b\]and separation \[d.\] \[\therefore \]      \[{{C}_{2}}=\frac{{{k}_{1}}{{\varepsilon }_{0}}\left( \frac{l}{2}\times b \right)}{d}+\frac{{{k}_{2}}{{\varepsilon }_{0}}\left( \frac{l}{2}\times b \right)}{d}\] \[\therefore \]      \[{{C}_{2}}=\frac{{{k}_{1}}{{\varepsilon }_{0}}(lb)}{2d}\,\,({{k}_{1}}+{{k}_{2}})\] Ratio of capacitance \[\frac{{{C}_{1}}}{{{C}_{2}}}=\frac{\frac{k{{\varepsilon }_{0}}(l\times b)}{d}}{\frac{{{\varepsilon }_{0}}(lb)}{2d}\,\,({{k}_{1}}+{{k}_{2}})}\] i.e.,                   \[\frac{{{C}_{1}}}{{{C}_{2}}}=\frac{k}{{{k}_{1}}+{{k}_{2}}}\] Or Gauss?s law Gauss?s law states that the total electric flux through a closed surface is equal to \[1\,/\,{{\varepsilon }_{0}}\] times the magnitude of the charged enclosed by it. \[\phi =\frac{q}{{{\varepsilon }_{0}}}\] Here, \[{{\varepsilon }_{0}}\] is the absolute permittivity of the free space and \[q\] is the total charge enclosed. \[\phi =\oint{E.ds=\frac{q}{{{\varepsilon }_{0}}}}\] (i) Magnitude of E at left face \[{{E}_{1}}=\alpha {{a}^{1/2}}\] Flux through left face \[{{\phi }_{1}}={{E}_{1}}.d{{s}_{1}}\]                                     \[=-\,\alpha {{a}^{1/2}}\,\text{\hat{i}}\times {{(0.1)}^{2}}(-\hat{i})\]                                     \[=-\,\alpha {{a}^{2}}\times 0.01\] Magnitude of \[E\] at right face, \[{{E}_{2}}=\alpha \,{{(2a)}^{1/2}}\] Flux through right face \[{{\phi }_{2}}={{E}_{2}}.d{{s}_{2}}\] \[=\alpha \,{{(2a)}^{1/2}}\,\text{\hat{i}}\text{.(0}\text{.1}{{\text{)}}^{2}}(\hat{i})\] \[=\alpha \,{{(2a)}^{1/2}}\times 0.01\] \[\therefore \] Total flux \[{{\phi }_{s}}={{\phi }_{1}}+{{\phi }_{2}}\]             \[=-\,\alpha {{a}^{1/2}}\times 0.01+\alpha \,{{(2a)}^{1/2}}\times 0.01\]             \[=800\sqrt{0.1}\times 0.01\,\,(\sqrt{2}-1)\,\,(a=0.1\,m)\] \[=1.04734\,W\approx 1.05\,N\text{-}{{m}^{2}}{{C}^{-1}}\]             (ii) \[{{\phi }_{s}}=\frac{q}{{{\varepsilon }_{0}}}\] (According to Gauss law)             \[\therefore \] \[q={{\phi }_{s}}\times {{\varepsilon }_{0}}\]             Where, \[{{\varepsilon }_{0}}=\]permittivity of free space                          \[=8.854\times {{10}^{-12}}\,{{N}^{-1}}{{C}^{2}}{{m}^{-2}}\]             \[\therefore \]      \[q=1.04734\times 8.854\times {{10}^{-12}}\]                          \[=9.2\times {{10}^{-12}}C\]