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Standard XI

Physics

Gauss's Law

The electric ...

Question

The electric flux for Gaussian surface A that enclose the charged particles in free space is $(g i v e n q_{1} = - 14 n C, q_{2} = 78.85 n C, q_{3} = - 56 n C)$

10^{3} N m^{2} c^{- 1}

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10^{3} C N^{- 1} m^{- 2}

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

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

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Solution

The correct option is A $10^{3} N m^{2} c^{- 1}$

Flux is due to charges enclosed per $ϵ_{0}$
$∴$ Total flux $= (- 14 + 78.85 - 56) n C / ϵ_{0}$
$= 8.85 \times 10^{- 9} C \times \frac{4 π}{4 π ϵ_{0}} = 8.85 \times 10^{- 9} \times 9 \times 10^{9} \times 4 π$
$= 1000.4 N m^{2} / C i . . e . 1000 N m^{2} C^{- 1}$

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Similar questions

The electric flux for Gaussian surface A that enclose the charged particles in free space is

(g i v e n q_{1} = - 14 n C, q_{2} = 78.85 n C, q_{3} = - 56 n C)

Q. Assertion :Four point charges

q_{1}

q_{2}

q_{3}

and

q_{4}

are as shown in figure. The flux over the shown Gaussian surface depends only on charges

q_{1}

and

q_{2}

. Reason: Electric field at all points on Gaussian surface depends only charges

q_{1}

and

q_{2}

Q. Assertion

(A)

: Four point charges

q_{1}, q_{2}, q_{3}

and

q_{4}

are as shown in figure. The flux over the shown gaussian surface depends only on charges

q_{1}

and

q_{2}

Reason

(R)

: Electric field at all points on gaussian surface is depends only on charges

q_{1}

and

q_{2}

Select the most appropriate answer from the options given below

In the figure shown, the particles have charges $q_{1}$ = $- q_{2}$ = 300 nC and $q_{3}$ = $- q_{4}$ =200 nC, and distance a = 5.0 cm. What are the (A) magnitude and (B) angle (relative to the +x direction) of the net force on particle 3?

Q. The electric flux through a closed surface is found to be

780 N - m^{2}

/C Approximately what quantity of charge is enclosed by the surface in nC?

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The electric flux for Gaussian surface A that enclose the charged particles in free space is (given q1=−14 nC, q2=78.85 nC, q3=−56 nC)

The correct option is A 103 Nm2 c−1 Flux is due to charges enclosed per ϵ0 ∴ Total flux =(−14+78.85−56)nC/ϵ0 =8.85×10−9C×4π4πϵ0=8.85×10−9×9×109×4π =1000.4 Nm2/C i..e. 1000 Nm2C−1

The electric flux for Gaussian surface A that enclose the charged particles in free space is $(g i v e n q_{1} = - 14 n C, q_{2} = 78.85 n C, q_{3} = - 56 n C)$