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Electric Flux Through a Cube |
Net electric flux through a cube is the sum of fluxes through its six faces. Consider a cube as shown in figure, having sides of length L = 10.0 cm. The electric field is uniform, has a magnitude \[E=4.00\times {{10}^{3}}N{{C}^{-1}}\]and is parallel to the xy plane at an angle of \[37{}^\circ \]measured from the 4-x-axis towards the +y-axis. |
A) \[-24\,N\,{{m}^{2}}{{C}^{-1}}\]
B) \[24\,N\,{{m}^{2}}\,{{C}^{-1}}\]
C) \[32\,N\,{{m}^{2}}\,\,{{C}^{-1}}\]
D) \[-32\,N{{m}^{2}}\,{{C}^{-1}}\]
Correct Answer: D
Solution :
Electric flux, \[\phi =\overrightarrow{E}\,.\,\overrightarrow{A}=EA\,\cos \theta \], where \[\overrightarrow{A}=A\hat{n}\] For electric flux passing through \[{{S}_{6}}\], \[{{\hat{n}}_{{{S}_{6}}}}=-\hat{i}\] (Back) \[\therefore \,\,{{\phi }_{{{S}_{6}}}}=-\left( 4\times {{10}^{3}}\,N\,{{C}^{-1}} \right){{\left( 0.1\,m \right)}^{2}}\,\cos \,37{}^\circ \] \[=-32\,N\,{{m}^{2}}{{C}^{-1}}\]You need to login to perform this action.
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