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1. Electric Charges and Fields
medium
Expression for an electric field is given by $\vec{E}=4000 x^2 \hat{i} \frac{V}{m}$. The electric flux through the cube of side $20\,cm$ when placed in electric field (as shown in the figure) is $.........V cm$.
A
$640$
B
$689$
C
$652$
D
$258$
(JEE MAIN-2023)
Solution
Flux $=\overrightarrow{ E } \cdot \overrightarrow{ A }$
$=4000(0 \cdot 2)^2 \frac{ V }{ m } \cdot(0 \cdot 2)^2 m ^2$
$=4000 \times 16 \times 10^{-4}\,Vm$
$=640\,V\,cm$
Standard 12
Physics
