Question

The electric field components in the figure are $E_x=\alpha x^{1/2},E_y=0,E_z=0$ where $\alpha =800N/m^2$. If a = 0.1 m is the side of cube then the charge within the cube is:

• A. $9.3\times {10}^{-12}C$
• B. $6\times {10}^{-12}C$
• C. $2.5\times {10}^{-12}C$
• D. Zero

Answer 1

The correct option is A $9.3\times {10}^{-12}C$

Flux at surface $x=a{\varphi }_a=a^2\times \left(\alpha a^{\frac{1}{2}}\right)$

(as area element is outside and E is directed inside)

$\therefore {\varphi }_a=-\alpha a^{2.5}$

Flux at the surface ${\varphi }_{2a}=a^2\times \left(\alpha \right(2a{)}^{\frac{1}{2}})$

${\varphi }_{2a}=\alpha \sqrt{2}\cdot a^{2.5}$

total flux $=\alpha a^{2.5}(\sqrt{2}-1)$

we know $\varphi =\frac{q}{{\epsilon }_0}\Rightarrow q=\alpha a^{2.5}(\sqrt{2}-1)\times 8\cdot 85\times {10}^{-12}C$

$\therefore q=800\times (0\cdot 1{)}^2\times \sqrt{0\cdot 1}(0\cdot 414)\times 8\cdot 85\times {10}^{-12}C$

$q=29.427\times \sqrt{0\cdot 1}\times {10}^{-12}$

$\therefore q=9.303\times {10}^{-12}C.$