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The electric field in a region is radially outward with magnitude E=A r. the charge contained in a sphere of radius a centred at the osigion. Take A=100 \mathrm{~V} / \mathrm{m}^{2}$. and $a=20 \mathrm{~cm}.

Option: 1

9 \times 10^{-12} \mathrm{c}


Option: 2

8 \times 10^{-13} \mathrm{C}


Option: 3

8.89 \times 10^{-11} \mathrm{C}


Option: 4

None


Answers (1)

best_answer

The flux of electric field is -

\mathrm{4=\oint E \cdot d s \cos \theta =A a\left(4 \pi a^{2}\right) }
                                       \mathrm{=4 \pi A a^{3} }

The charge contained in the sphere is, from gauss law -

\mathrm{Q_{\text {inside }} =\varepsilon_{0}\phi+=4 \pi \epsilon_{0} \mathrm{Aa}^{3}}
                \mathrm{=\left(\frac{1}{9 \times 10^{9}} \mathrm{c}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{2}\right)\left(100 \mathrm{vm}^{-2}\right)\left(0.20 \mathrm{~m}^{3}\right.)}
                \mathrm{=\frac{1}{9 \times 10^{9}} \times 10^{2} \times(0.20)^{3}}

\mathrm{Q_{\text {inside }} =8.89 \times 10^{-11} \mathrm{c} } Ans.
             
 

Posted by

Pankaj

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