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Expression for an electric field is given by \mathrm{\vec{E}=4000x^{2}\: \hat{i}\: \frac{v}{m}} . The electric flux through the cube of side \mathrm{20 cm} when placed in electric field (as shown in the figure) is _____ \mathrm{V cm}

Option: 1

640


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} & \vec{E} \perp \vec{A}, \phi_{\text {Top }}=\phi_{\text {Bottom }}=\phi_{\text {front }}=\phi_{\text {Back }}=0 \\ \end{aligned}

\begin{aligned} & \text { for } O C D G, x=0, E=0, \phi=0 \\ \end{aligned}

\begin{aligned} & \text { for } A B E F, x=0.2 \mathrm{~m} \\ \end{aligned}

\begin{aligned} & E=4000 \times(0.2)^2 \\ \end{aligned}

\begin{aligned} & E=160 \mathrm{~V} / \mathrm{m} \\ \end{aligned}

\begin{aligned} & \phi=E\left(a^2\right)=160 \mathrm{~V} / \mathrm{m} \times(0.2)^2 \mathrm{~m}^2 \\ \end{aligned}

\begin{aligned} & \phi=6.4 \mathrm{~V}-\mathrm{m} \\ \end{aligned}

\begin{aligned} & \phi=640 \mathrm{~V}-\mathrm{cm} \end{aligned}

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Shailly goel

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