Electric Charges And Fields Ques 29
29. The electric field in a certain region is acting radially outward and is given by $E=Aa$. $A$ charge contained in a sphere of radius ’ $a$ ’ centred at the origin of the field, will be given by[2015]
(a) $A \varepsilon_0 a^{2}$
(b) $4 \pi \varepsilon_0 Aa^{3}$
(c) $\varepsilon_0 Aa^{3}$
(d) $4 \pi \varepsilon_0 Aa^{2}$
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Answer:
Correct Answer: 29.(b)
Solution:
- (b) Net flux emitted from a spherical surface of radius a according to Gauss’s theorem
$\phi _{\text{net }}=\frac{q _{\text{in }}}{\varepsilon_0}$ or, $(A a)(4 \pi a^{2})=\frac{q _{\text{in }}}{\varepsilon_0}$
So, $q _{in}=4 \pi \varepsilon_0 A a^{3}$
Electric flux is equal to the product of an area element $A$ and the perpendicular component of electric field $E$ integrated over a surface. $\phi=$ E.A $\cos \theta$
BETA
