358316
\({{\rm{q}}_1},{q_2},{q_3}\,{\rm{and}}\,{q_4}\)are point charges located at points as shown in the figure and \(S\) is a spherical Gaussian surface of radius \(R\). Which of the following is true according to Gauss’s law?
From Gauss’s law \(\oint {\overline E .d\overline A } = \frac{{{q_{total}}}}{{{\varepsilon _o}}}\) Where \(\overline {\rm{E}} \) is the resultant electric field at a point by both inside and outside charges. \({q_{total}}\) is the net charge enclosed inside the gaussian surface. \(\overline E = {\overline E _1} + {\overline E _2} + {\overline E _3} + {\overline E _4}\) \(q = {q_1} + {q_2} + {q_3}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358317
A hollow cylinder has a charge \(q\) coulomb within it. If \(\phi \) is the electric flux associated with the curved surface \(B\), the flux linked with the plane surface \(A\) in units of volt-meter will be
358318
A point charge \( + Q\) is positioned at the centre of the base of a square pyramid as shown. The flux through one of the four identical upper faces of the pyramid is
1 \(\frac{Q}{{16{\varepsilon _0}}}\)
2 \(\frac{Q}{{4{\varepsilon _0}}}\)
3 \(\frac{Q}{{8{\varepsilon _0}}}\)
4 None of these
Explanation:
Flux going in pyramid \( = \frac{Q}{{2{\varepsilon _0}}}\) Which is divided equally among all 4 faces. \(\therefore \) Flux through one face \( = \frac{Q}{{8{\varepsilon _0}}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358319
Electric flux emanating through a surface element \(d s=5 \hat{i}\) placed in an electric field \(E=4 \hat{i}+4 \hat{j}+4 \hat{k}\) is
358320
If the electric flux entering and leaving a closed surface are \(6 \times {10^6}\,{\rm{and}}\,9 \times {10^6}\) S.I. units respectively, then the charge inside the surface of permittivity of free space \({\varepsilon _0}\) is
1 \({\varepsilon _0} \times {10^6}\)
2 \( - {\varepsilon _0} \times {10^6}\)
3 \( - 2{\varepsilon _0} \times {10^6}\)
4 \(3{\varepsilon _0} \times {10^6}\)
Explanation:
By Gauss law, we know that \(\phi = \frac{q}{{{\varepsilon _0}}}\) Here, Net electric flux, \(\phi = \phi {}_2 - {\phi _1}\) \( = 9 \times {10^6} - 6 \times {10^6} = \frac{q}{{{\varepsilon _0}}} \Rightarrow q = 3 \times {10^6} \times {\varepsilon _0}\)
358316
\({{\rm{q}}_1},{q_2},{q_3}\,{\rm{and}}\,{q_4}\)are point charges located at points as shown in the figure and \(S\) is a spherical Gaussian surface of radius \(R\). Which of the following is true according to Gauss’s law?
From Gauss’s law \(\oint {\overline E .d\overline A } = \frac{{{q_{total}}}}{{{\varepsilon _o}}}\) Where \(\overline {\rm{E}} \) is the resultant electric field at a point by both inside and outside charges. \({q_{total}}\) is the net charge enclosed inside the gaussian surface. \(\overline E = {\overline E _1} + {\overline E _2} + {\overline E _3} + {\overline E _4}\) \(q = {q_1} + {q_2} + {q_3}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358317
A hollow cylinder has a charge \(q\) coulomb within it. If \(\phi \) is the electric flux associated with the curved surface \(B\), the flux linked with the plane surface \(A\) in units of volt-meter will be
358318
A point charge \( + Q\) is positioned at the centre of the base of a square pyramid as shown. The flux through one of the four identical upper faces of the pyramid is
1 \(\frac{Q}{{16{\varepsilon _0}}}\)
2 \(\frac{Q}{{4{\varepsilon _0}}}\)
3 \(\frac{Q}{{8{\varepsilon _0}}}\)
4 None of these
Explanation:
Flux going in pyramid \( = \frac{Q}{{2{\varepsilon _0}}}\) Which is divided equally among all 4 faces. \(\therefore \) Flux through one face \( = \frac{Q}{{8{\varepsilon _0}}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358319
Electric flux emanating through a surface element \(d s=5 \hat{i}\) placed in an electric field \(E=4 \hat{i}+4 \hat{j}+4 \hat{k}\) is
358320
If the electric flux entering and leaving a closed surface are \(6 \times {10^6}\,{\rm{and}}\,9 \times {10^6}\) S.I. units respectively, then the charge inside the surface of permittivity of free space \({\varepsilon _0}\) is
1 \({\varepsilon _0} \times {10^6}\)
2 \( - {\varepsilon _0} \times {10^6}\)
3 \( - 2{\varepsilon _0} \times {10^6}\)
4 \(3{\varepsilon _0} \times {10^6}\)
Explanation:
By Gauss law, we know that \(\phi = \frac{q}{{{\varepsilon _0}}}\) Here, Net electric flux, \(\phi = \phi {}_2 - {\phi _1}\) \( = 9 \times {10^6} - 6 \times {10^6} = \frac{q}{{{\varepsilon _0}}} \Rightarrow q = 3 \times {10^6} \times {\varepsilon _0}\)
358316
\({{\rm{q}}_1},{q_2},{q_3}\,{\rm{and}}\,{q_4}\)are point charges located at points as shown in the figure and \(S\) is a spherical Gaussian surface of radius \(R\). Which of the following is true according to Gauss’s law?
From Gauss’s law \(\oint {\overline E .d\overline A } = \frac{{{q_{total}}}}{{{\varepsilon _o}}}\) Where \(\overline {\rm{E}} \) is the resultant electric field at a point by both inside and outside charges. \({q_{total}}\) is the net charge enclosed inside the gaussian surface. \(\overline E = {\overline E _1} + {\overline E _2} + {\overline E _3} + {\overline E _4}\) \(q = {q_1} + {q_2} + {q_3}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358317
A hollow cylinder has a charge \(q\) coulomb within it. If \(\phi \) is the electric flux associated with the curved surface \(B\), the flux linked with the plane surface \(A\) in units of volt-meter will be
358318
A point charge \( + Q\) is positioned at the centre of the base of a square pyramid as shown. The flux through one of the four identical upper faces of the pyramid is
1 \(\frac{Q}{{16{\varepsilon _0}}}\)
2 \(\frac{Q}{{4{\varepsilon _0}}}\)
3 \(\frac{Q}{{8{\varepsilon _0}}}\)
4 None of these
Explanation:
Flux going in pyramid \( = \frac{Q}{{2{\varepsilon _0}}}\) Which is divided equally among all 4 faces. \(\therefore \) Flux through one face \( = \frac{Q}{{8{\varepsilon _0}}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358319
Electric flux emanating through a surface element \(d s=5 \hat{i}\) placed in an electric field \(E=4 \hat{i}+4 \hat{j}+4 \hat{k}\) is
358320
If the electric flux entering and leaving a closed surface are \(6 \times {10^6}\,{\rm{and}}\,9 \times {10^6}\) S.I. units respectively, then the charge inside the surface of permittivity of free space \({\varepsilon _0}\) is
1 \({\varepsilon _0} \times {10^6}\)
2 \( - {\varepsilon _0} \times {10^6}\)
3 \( - 2{\varepsilon _0} \times {10^6}\)
4 \(3{\varepsilon _0} \times {10^6}\)
Explanation:
By Gauss law, we know that \(\phi = \frac{q}{{{\varepsilon _0}}}\) Here, Net electric flux, \(\phi = \phi {}_2 - {\phi _1}\) \( = 9 \times {10^6} - 6 \times {10^6} = \frac{q}{{{\varepsilon _0}}} \Rightarrow q = 3 \times {10^6} \times {\varepsilon _0}\)
358316
\({{\rm{q}}_1},{q_2},{q_3}\,{\rm{and}}\,{q_4}\)are point charges located at points as shown in the figure and \(S\) is a spherical Gaussian surface of radius \(R\). Which of the following is true according to Gauss’s law?
From Gauss’s law \(\oint {\overline E .d\overline A } = \frac{{{q_{total}}}}{{{\varepsilon _o}}}\) Where \(\overline {\rm{E}} \) is the resultant electric field at a point by both inside and outside charges. \({q_{total}}\) is the net charge enclosed inside the gaussian surface. \(\overline E = {\overline E _1} + {\overline E _2} + {\overline E _3} + {\overline E _4}\) \(q = {q_1} + {q_2} + {q_3}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358317
A hollow cylinder has a charge \(q\) coulomb within it. If \(\phi \) is the electric flux associated with the curved surface \(B\), the flux linked with the plane surface \(A\) in units of volt-meter will be
358318
A point charge \( + Q\) is positioned at the centre of the base of a square pyramid as shown. The flux through one of the four identical upper faces of the pyramid is
1 \(\frac{Q}{{16{\varepsilon _0}}}\)
2 \(\frac{Q}{{4{\varepsilon _0}}}\)
3 \(\frac{Q}{{8{\varepsilon _0}}}\)
4 None of these
Explanation:
Flux going in pyramid \( = \frac{Q}{{2{\varepsilon _0}}}\) Which is divided equally among all 4 faces. \(\therefore \) Flux through one face \( = \frac{Q}{{8{\varepsilon _0}}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358319
Electric flux emanating through a surface element \(d s=5 \hat{i}\) placed in an electric field \(E=4 \hat{i}+4 \hat{j}+4 \hat{k}\) is
358320
If the electric flux entering and leaving a closed surface are \(6 \times {10^6}\,{\rm{and}}\,9 \times {10^6}\) S.I. units respectively, then the charge inside the surface of permittivity of free space \({\varepsilon _0}\) is
1 \({\varepsilon _0} \times {10^6}\)
2 \( - {\varepsilon _0} \times {10^6}\)
3 \( - 2{\varepsilon _0} \times {10^6}\)
4 \(3{\varepsilon _0} \times {10^6}\)
Explanation:
By Gauss law, we know that \(\phi = \frac{q}{{{\varepsilon _0}}}\) Here, Net electric flux, \(\phi = \phi {}_2 - {\phi _1}\) \( = 9 \times {10^6} - 6 \times {10^6} = \frac{q}{{{\varepsilon _0}}} \Rightarrow q = 3 \times {10^6} \times {\varepsilon _0}\)
358316
\({{\rm{q}}_1},{q_2},{q_3}\,{\rm{and}}\,{q_4}\)are point charges located at points as shown in the figure and \(S\) is a spherical Gaussian surface of radius \(R\). Which of the following is true according to Gauss’s law?
From Gauss’s law \(\oint {\overline E .d\overline A } = \frac{{{q_{total}}}}{{{\varepsilon _o}}}\) Where \(\overline {\rm{E}} \) is the resultant electric field at a point by both inside and outside charges. \({q_{total}}\) is the net charge enclosed inside the gaussian surface. \(\overline E = {\overline E _1} + {\overline E _2} + {\overline E _3} + {\overline E _4}\) \(q = {q_1} + {q_2} + {q_3}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358317
A hollow cylinder has a charge \(q\) coulomb within it. If \(\phi \) is the electric flux associated with the curved surface \(B\), the flux linked with the plane surface \(A\) in units of volt-meter will be
358318
A point charge \( + Q\) is positioned at the centre of the base of a square pyramid as shown. The flux through one of the four identical upper faces of the pyramid is
1 \(\frac{Q}{{16{\varepsilon _0}}}\)
2 \(\frac{Q}{{4{\varepsilon _0}}}\)
3 \(\frac{Q}{{8{\varepsilon _0}}}\)
4 None of these
Explanation:
Flux going in pyramid \( = \frac{Q}{{2{\varepsilon _0}}}\) Which is divided equally among all 4 faces. \(\therefore \) Flux through one face \( = \frac{Q}{{8{\varepsilon _0}}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358319
Electric flux emanating through a surface element \(d s=5 \hat{i}\) placed in an electric field \(E=4 \hat{i}+4 \hat{j}+4 \hat{k}\) is
358320
If the electric flux entering and leaving a closed surface are \(6 \times {10^6}\,{\rm{and}}\,9 \times {10^6}\) S.I. units respectively, then the charge inside the surface of permittivity of free space \({\varepsilon _0}\) is
1 \({\varepsilon _0} \times {10^6}\)
2 \( - {\varepsilon _0} \times {10^6}\)
3 \( - 2{\varepsilon _0} \times {10^6}\)
4 \(3{\varepsilon _0} \times {10^6}\)
Explanation:
By Gauss law, we know that \(\phi = \frac{q}{{{\varepsilon _0}}}\) Here, Net electric flux, \(\phi = \phi {}_2 - {\phi _1}\) \( = 9 \times {10^6} - 6 \times {10^6} = \frac{q}{{{\varepsilon _0}}} \Rightarrow q = 3 \times {10^6} \times {\varepsilon _0}\)
