Linear charge density of wire is 8.85,μ C/m . Radius and height of cylinder are 3,m and 4,m . Then

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1. Electric Charges and Fields

medium

Linear charge density of wire is $8.85\,\mu C/m$ . Radius and height of the cylinder are $3\,m$ and $4\,m$ . Then find the flux passing through the cylinder

$5 \times 10^6\, volt-m$

$3 \times 10^6\, volt-m$

$4 \times 10^6\, volt-m$

None

Solution

$\phi = \frac{{\lambda {\ell _{{\rm{enclosed }}}}}}{{{ \in _0}}} = \frac{{8.85 \times {{10}^{ – 6}} \times 5}}{{8.85 \times {{10}^{ – 12}}}} = 5 \times {10^6}{\mkern 1mu} \,{\rm{V}} – {\rm{m}}$

Standard 12

Physics

For a given surface the Gauss's law is stated as $\oint {E \cdot ds} = 0$. From this we can conclude that

easy

Figure shows four charges $q_1, q_2, q_3$ and $q_4$ fixed in space. Then the total flux of electric field through a closed surface $S$, due to all charges $q_1, q_2, q_3$ and $q_4$ is

medium

Gauss’s law is true only if force due to a charge varies as

easy

Draw electric field lines when two positive charges are near.

medium

Three charges $q_1 = 1\,\mu c, q_2 = 2\,\mu c$ and $q_3 = -3\,\mu c$ and four surfaces $S_1, S_2 ,S_3$ and $S_4$ are shown in figure. The flux emerging through surface $S_2$ in $N-m^2/C$ is

medium

Linear charge density of wire is $8.85\,\mu C/m$ . Radius and height of the cylinder are $3\,m$ and $4\,m$ . Then find the flux passing through the cylinder

Solution

Similar Questions

For a given surface the Gauss's law is stated as $\oint {E \cdot ds} = 0$. From this we can conclude that

Figure shows four charges $q_1, q_2, q_3$ and $q_4$ fixed in space. Then the total flux of electric field through a closed surface $S$, due to all charges $q_1, q_2, q_3$ and $q_4$ is

Gauss’s law is true only if force due to a charge varies as

Draw electric field lines when two positive charges are near.

Three charges $q_1 = 1\,\mu c, q_2 = 2\,\mu c$ and $q_3 = -3\,\mu c$ and four surfaces $S_1, S_2 ,S_3$ and $S_4$ are shown in figure. The flux emerging through surface $S_2$ in $N-m^2/C$ is

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