(i) The flux passing through \[{{A}_{1}}\] is given by \[\frac{Q}{4{{\varepsilon }_{0}}}\]. |
(ii) The flux passing through \[{{A}_{2}}\] is given by \[\frac{Q}{9{{\varepsilon }_{0}}}\](iii) The flux passing through \[{{A}_{2}}\] within the area equal to \[{{A}_{1}}\] is given by\[\frac{Q}{16{{\varepsilon }_{0}}}\]. |
(iv) The flux passing through \[{{A}_{3}}\] within the area equal to \[{{A}_{2}}\] is given by \[\frac{Q}{4{{\varepsilon }_{0}}}\] |
The correct statements are |
A) (i) and (iii)
B) (ii) and (iii)
C) (i) and (iv)
D) (iii) and (iv)
Correct Answer: A
Solution :
[a] We have \[{{A}_{1}}=4\pi {{r}^{2}}\] \[{{A}_{2}}=4\pi {{(2r)}^{2}}=4{{A}_{1}}\] \[{{A}_{3}}=4\pi {{(3r)}^{2}}=9{{A}_{1}}=\frac{9}{4}{{A}_{2}}\] \[{{\phi }_{1}}={{\phi }_{2}}={{\phi }_{3}}\] through same solid angle \[{{\phi }_{1}}=\int{EdA=\int{\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{{{r}^{2}}}}dA=\frac{Q}{4\pi {{\varepsilon }_{0}}}\Omega =\frac{Q}{4\pi {{\varepsilon }_{0}}}\pi =\frac{Q}{4{{\varepsilon }_{0}}}}\] Flux passing through \[{{A}_{1}}\] within area \[{{A}_{1}}=\frac{1}{4}\frac{Q}{4{{\varepsilon }_{0}}}=\frac{Q}{16{{\varepsilon }_{0}}}\] Flux passing through \[{{A}_{2}}\] within area \[{{A}_{2}}=\frac{1}{9}\frac{4Q}{4{{\varepsilon }_{0}}}=\frac{Q}{9{{\varepsilon }_{0}}}\]You need to login to perform this action.
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