From gauss law \(\phi=\dfrac{q}{\varepsilon_{0}}\) This is the net flux coming out of the cube. Since a cube has 6 sides. So electric flux through any face is \(\phi_{1}=\dfrac{q}{6 \varepsilon_{0}}=\dfrac{4 \pi q}{q\left(4 \pi \varepsilon_{0}\right)}\) For two faces, flux \(=2 \phi_{1}=\dfrac{4 \pi q}{3\left(4 \pi \varepsilon_{0}\right)}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358327
A cube of side \(l\) is placed in a uniform field \(E\), where \(E=E_{i}\). The net electric flux through the cube is
1 zero
2 \(l^{2} E\)
3 \(4 l^{2} E\)
4 \(6 l^{2} E\)
Explanation:
Incoming flux is equal to outgoing flux for a cube placed in a uniform electric field. So net flux through cube will be zero.
PHXII01:ELECTRIC CHARGES AND FIELDS
358328
Shown below is a distribution of charges. The flux of electric field due to these charges through the surface is
1 \(2q/{\varepsilon _0}\)
2 \(q/{\varepsilon _0}\)
3 \(3q/{\varepsilon _0}\)
4 Zero
Explanation:
The total flux is \(\phi = \frac{{{q_{enc}}}}{{{\varepsilon _0}}} = \frac{1}{{{\varepsilon _0}}}(2q)\)
From gauss law \(\phi=\dfrac{q}{\varepsilon_{0}}\) This is the net flux coming out of the cube. Since a cube has 6 sides. So electric flux through any face is \(\phi_{1}=\dfrac{q}{6 \varepsilon_{0}}=\dfrac{4 \pi q}{q\left(4 \pi \varepsilon_{0}\right)}\) For two faces, flux \(=2 \phi_{1}=\dfrac{4 \pi q}{3\left(4 \pi \varepsilon_{0}\right)}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358327
A cube of side \(l\) is placed in a uniform field \(E\), where \(E=E_{i}\). The net electric flux through the cube is
1 zero
2 \(l^{2} E\)
3 \(4 l^{2} E\)
4 \(6 l^{2} E\)
Explanation:
Incoming flux is equal to outgoing flux for a cube placed in a uniform electric field. So net flux through cube will be zero.
PHXII01:ELECTRIC CHARGES AND FIELDS
358328
Shown below is a distribution of charges. The flux of electric field due to these charges through the surface is
1 \(2q/{\varepsilon _0}\)
2 \(q/{\varepsilon _0}\)
3 \(3q/{\varepsilon _0}\)
4 Zero
Explanation:
The total flux is \(\phi = \frac{{{q_{enc}}}}{{{\varepsilon _0}}} = \frac{1}{{{\varepsilon _0}}}(2q)\)
From gauss law \(\phi=\dfrac{q}{\varepsilon_{0}}\) This is the net flux coming out of the cube. Since a cube has 6 sides. So electric flux through any face is \(\phi_{1}=\dfrac{q}{6 \varepsilon_{0}}=\dfrac{4 \pi q}{q\left(4 \pi \varepsilon_{0}\right)}\) For two faces, flux \(=2 \phi_{1}=\dfrac{4 \pi q}{3\left(4 \pi \varepsilon_{0}\right)}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358327
A cube of side \(l\) is placed in a uniform field \(E\), where \(E=E_{i}\). The net electric flux through the cube is
1 zero
2 \(l^{2} E\)
3 \(4 l^{2} E\)
4 \(6 l^{2} E\)
Explanation:
Incoming flux is equal to outgoing flux for a cube placed in a uniform electric field. So net flux through cube will be zero.
PHXII01:ELECTRIC CHARGES AND FIELDS
358328
Shown below is a distribution of charges. The flux of electric field due to these charges through the surface is
1 \(2q/{\varepsilon _0}\)
2 \(q/{\varepsilon _0}\)
3 \(3q/{\varepsilon _0}\)
4 Zero
Explanation:
The total flux is \(\phi = \frac{{{q_{enc}}}}{{{\varepsilon _0}}} = \frac{1}{{{\varepsilon _0}}}(2q)\)
From gauss law \(\phi=\dfrac{q}{\varepsilon_{0}}\) This is the net flux coming out of the cube. Since a cube has 6 sides. So electric flux through any face is \(\phi_{1}=\dfrac{q}{6 \varepsilon_{0}}=\dfrac{4 \pi q}{q\left(4 \pi \varepsilon_{0}\right)}\) For two faces, flux \(=2 \phi_{1}=\dfrac{4 \pi q}{3\left(4 \pi \varepsilon_{0}\right)}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358327
A cube of side \(l\) is placed in a uniform field \(E\), where \(E=E_{i}\). The net electric flux through the cube is
1 zero
2 \(l^{2} E\)
3 \(4 l^{2} E\)
4 \(6 l^{2} E\)
Explanation:
Incoming flux is equal to outgoing flux for a cube placed in a uniform electric field. So net flux through cube will be zero.
PHXII01:ELECTRIC CHARGES AND FIELDS
358328
Shown below is a distribution of charges. The flux of electric field due to these charges through the surface is
1 \(2q/{\varepsilon _0}\)
2 \(q/{\varepsilon _0}\)
3 \(3q/{\varepsilon _0}\)
4 Zero
Explanation:
The total flux is \(\phi = \frac{{{q_{enc}}}}{{{\varepsilon _0}}} = \frac{1}{{{\varepsilon _0}}}(2q)\)
