Electric flux for Gaussian surface A that enclose charged particles in free space is (given q₁ = -

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1. Electric Charges and Fields

medium

The electric flux for Gaussian surface A that enclose the charged particles in free space is (given $q_1$ = $-14\, nC$, $q_2$ = $78.85\, nC$, $q_3$ = $-56 \,nC$)

$10^3\,\,N{m^2}{C^{ - 1}}$

$10^3\,\,C{N^{-1}}{m^{ - 2}}$

$6.32 \times 10^3\,\,N{m^2}{C^{ - 1}}$

$6.32 \times 10^3\,\,C{N^{-1}}{m^{ - 2}}$

Solution

(a) Flux is due to charges enclosed per ${\varepsilon _0}$
Total flux = $( – 14 + 78.85 – 56)\,nC/{\varepsilon _0}$
$ = 8.85 \times {10^{ – 9}}\,C \times \frac{{4\pi }}{{4\pi {\varepsilon _0}}} = 8.85 \times {10^{ – 9}} \times 9 \times {10^9} \times 4\pi $
$ = 1000.4\,N{m^2}/C$ i..e. $1000\,\,N{m^2}{C^{ – 1}}$

Standard 12

Physics

Which among the curves shown in Figureb cannot possibly represent electrostatic field lines?

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Draw electric field by positive charge.

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hard

(JEE MAIN-2021)

The figure shows a hollow hemisphere of radius $R$ in which two charges $3q$ and $5q$ are placed symmetrically about the centre $O$ on the planar surface. The electric flux over the curved surface is

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easy

(AIPMT-2006)

The electric flux for Gaussian surface A that enclose the charged particles in free space is (given $q_1$ = $-14\, nC$, $q_2$ = $78.85\, nC$, $q_3$ = $-56 \,nC$)

Solution

Similar Questions

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