358291
A circular disc of radius ' \(r\) ' is placed along the plane of paper. A uniform electric field \(\vec{E}\) is also present in the plane of paper. What amount of electric flux is associated with it?
1 \(E \pi r^{2}\)
2 zero
3 \(2 E \pi r\)
4 \(\dfrac{\pi r^{2}}{E}\)
Explanation:
Electric lines of forces are not passing through it so flux is zero and electric lines of force are perpendicular to the areal plane of disc so according to the relation it will again be zero.
PHXII01:ELECTRIC CHARGES AND FIELDS
358292
Assertion : The electric flux emanating out and entering a closed surface ( in vacuum) are \(8 \times 10^{3}\) and \(2 \times {10^3}\,V\,m\) respectively. The charge enclosed by the surface is \(0.053\,\mu C\). Reason : Gauss's theorem is not relevant here.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(\phi=Q_{\text {enclosed }} / \in_{0}\) \({Q_{enclosed{\rm{ }}}} = {\smallint _0}\phi \) \( = 8.85 \times {10^{ - 12}}\left[ {8 \times {{10}^3} - 2 \times {{10}^3}} \right]\) \( = 53.10 \times {10^{ - 9}}C\) \( = 0.053\mu C\) Gauss law is very much applicable as some net electric charge is in a closed surface. So correct option is (3).
PHXII01:ELECTRIC CHARGES AND FIELDS
358293
Five electric dipoles of charge ' \(q\) ' each are placed inside the shell. What will be the amount of electric flux associated with the shell?
1 \(5 q \varepsilon_{0}\)
2 \(\dfrac{10 q}{\varepsilon_{0}}\)
3 Zero
4 \(\dfrac{5 q}{\varepsilon_{0}}\)
Explanation:
Total charge into the shell is zero because a dipole composed of negative and positive charge. So, \(q_{n e t}=5(-q+q)=0\) \(\phi=\dfrac{q}{\varepsilon_{0}}=\text { zero . }\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358294
If \(\oint\limits_s {\vec E} \cdot \overrightarrow {dS} = 0\) over a surface, then:
1 The magnitude of electric field on the surface is constant.
2 All the charges must necessarily be inside the surface
3 The electric field inside the surface is necessarily uniform
4 The number of flux lines entering the surface must be equal to the number of flux lines leaving it.
Explanation:
For diagram shown: \(\begin{aligned}& \phi=-E d s+E d s \\& \Rightarrow \phi=0 \\& \Rightarrow \phi_{\text {in }}=\phi_{\text {out }}\end{aligned}\) Hence, number of field lines entering is equal to number of field lines leaving. Correct option is (4).
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII01:ELECTRIC CHARGES AND FIELDS
358291
A circular disc of radius ' \(r\) ' is placed along the plane of paper. A uniform electric field \(\vec{E}\) is also present in the plane of paper. What amount of electric flux is associated with it?
1 \(E \pi r^{2}\)
2 zero
3 \(2 E \pi r\)
4 \(\dfrac{\pi r^{2}}{E}\)
Explanation:
Electric lines of forces are not passing through it so flux is zero and electric lines of force are perpendicular to the areal plane of disc so according to the relation it will again be zero.
PHXII01:ELECTRIC CHARGES AND FIELDS
358292
Assertion : The electric flux emanating out and entering a closed surface ( in vacuum) are \(8 \times 10^{3}\) and \(2 \times {10^3}\,V\,m\) respectively. The charge enclosed by the surface is \(0.053\,\mu C\). Reason : Gauss's theorem is not relevant here.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(\phi=Q_{\text {enclosed }} / \in_{0}\) \({Q_{enclosed{\rm{ }}}} = {\smallint _0}\phi \) \( = 8.85 \times {10^{ - 12}}\left[ {8 \times {{10}^3} - 2 \times {{10}^3}} \right]\) \( = 53.10 \times {10^{ - 9}}C\) \( = 0.053\mu C\) Gauss law is very much applicable as some net electric charge is in a closed surface. So correct option is (3).
PHXII01:ELECTRIC CHARGES AND FIELDS
358293
Five electric dipoles of charge ' \(q\) ' each are placed inside the shell. What will be the amount of electric flux associated with the shell?
1 \(5 q \varepsilon_{0}\)
2 \(\dfrac{10 q}{\varepsilon_{0}}\)
3 Zero
4 \(\dfrac{5 q}{\varepsilon_{0}}\)
Explanation:
Total charge into the shell is zero because a dipole composed of negative and positive charge. So, \(q_{n e t}=5(-q+q)=0\) \(\phi=\dfrac{q}{\varepsilon_{0}}=\text { zero . }\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358294
If \(\oint\limits_s {\vec E} \cdot \overrightarrow {dS} = 0\) over a surface, then:
1 The magnitude of electric field on the surface is constant.
2 All the charges must necessarily be inside the surface
3 The electric field inside the surface is necessarily uniform
4 The number of flux lines entering the surface must be equal to the number of flux lines leaving it.
Explanation:
For diagram shown: \(\begin{aligned}& \phi=-E d s+E d s \\& \Rightarrow \phi=0 \\& \Rightarrow \phi_{\text {in }}=\phi_{\text {out }}\end{aligned}\) Hence, number of field lines entering is equal to number of field lines leaving. Correct option is (4).
358291
A circular disc of radius ' \(r\) ' is placed along the plane of paper. A uniform electric field \(\vec{E}\) is also present in the plane of paper. What amount of electric flux is associated with it?
1 \(E \pi r^{2}\)
2 zero
3 \(2 E \pi r\)
4 \(\dfrac{\pi r^{2}}{E}\)
Explanation:
Electric lines of forces are not passing through it so flux is zero and electric lines of force are perpendicular to the areal plane of disc so according to the relation it will again be zero.
PHXII01:ELECTRIC CHARGES AND FIELDS
358292
Assertion : The electric flux emanating out and entering a closed surface ( in vacuum) are \(8 \times 10^{3}\) and \(2 \times {10^3}\,V\,m\) respectively. The charge enclosed by the surface is \(0.053\,\mu C\). Reason : Gauss's theorem is not relevant here.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(\phi=Q_{\text {enclosed }} / \in_{0}\) \({Q_{enclosed{\rm{ }}}} = {\smallint _0}\phi \) \( = 8.85 \times {10^{ - 12}}\left[ {8 \times {{10}^3} - 2 \times {{10}^3}} \right]\) \( = 53.10 \times {10^{ - 9}}C\) \( = 0.053\mu C\) Gauss law is very much applicable as some net electric charge is in a closed surface. So correct option is (3).
PHXII01:ELECTRIC CHARGES AND FIELDS
358293
Five electric dipoles of charge ' \(q\) ' each are placed inside the shell. What will be the amount of electric flux associated with the shell?
1 \(5 q \varepsilon_{0}\)
2 \(\dfrac{10 q}{\varepsilon_{0}}\)
3 Zero
4 \(\dfrac{5 q}{\varepsilon_{0}}\)
Explanation:
Total charge into the shell is zero because a dipole composed of negative and positive charge. So, \(q_{n e t}=5(-q+q)=0\) \(\phi=\dfrac{q}{\varepsilon_{0}}=\text { zero . }\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358294
If \(\oint\limits_s {\vec E} \cdot \overrightarrow {dS} = 0\) over a surface, then:
1 The magnitude of electric field on the surface is constant.
2 All the charges must necessarily be inside the surface
3 The electric field inside the surface is necessarily uniform
4 The number of flux lines entering the surface must be equal to the number of flux lines leaving it.
Explanation:
For diagram shown: \(\begin{aligned}& \phi=-E d s+E d s \\& \Rightarrow \phi=0 \\& \Rightarrow \phi_{\text {in }}=\phi_{\text {out }}\end{aligned}\) Hence, number of field lines entering is equal to number of field lines leaving. Correct option is (4).
358291
A circular disc of radius ' \(r\) ' is placed along the plane of paper. A uniform electric field \(\vec{E}\) is also present in the plane of paper. What amount of electric flux is associated with it?
1 \(E \pi r^{2}\)
2 zero
3 \(2 E \pi r\)
4 \(\dfrac{\pi r^{2}}{E}\)
Explanation:
Electric lines of forces are not passing through it so flux is zero and electric lines of force are perpendicular to the areal plane of disc so according to the relation it will again be zero.
PHXII01:ELECTRIC CHARGES AND FIELDS
358292
Assertion : The electric flux emanating out and entering a closed surface ( in vacuum) are \(8 \times 10^{3}\) and \(2 \times {10^3}\,V\,m\) respectively. The charge enclosed by the surface is \(0.053\,\mu C\). Reason : Gauss's theorem is not relevant here.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(\phi=Q_{\text {enclosed }} / \in_{0}\) \({Q_{enclosed{\rm{ }}}} = {\smallint _0}\phi \) \( = 8.85 \times {10^{ - 12}}\left[ {8 \times {{10}^3} - 2 \times {{10}^3}} \right]\) \( = 53.10 \times {10^{ - 9}}C\) \( = 0.053\mu C\) Gauss law is very much applicable as some net electric charge is in a closed surface. So correct option is (3).
PHXII01:ELECTRIC CHARGES AND FIELDS
358293
Five electric dipoles of charge ' \(q\) ' each are placed inside the shell. What will be the amount of electric flux associated with the shell?
1 \(5 q \varepsilon_{0}\)
2 \(\dfrac{10 q}{\varepsilon_{0}}\)
3 Zero
4 \(\dfrac{5 q}{\varepsilon_{0}}\)
Explanation:
Total charge into the shell is zero because a dipole composed of negative and positive charge. So, \(q_{n e t}=5(-q+q)=0\) \(\phi=\dfrac{q}{\varepsilon_{0}}=\text { zero . }\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358294
If \(\oint\limits_s {\vec E} \cdot \overrightarrow {dS} = 0\) over a surface, then:
1 The magnitude of electric field on the surface is constant.
2 All the charges must necessarily be inside the surface
3 The electric field inside the surface is necessarily uniform
4 The number of flux lines entering the surface must be equal to the number of flux lines leaving it.
Explanation:
For diagram shown: \(\begin{aligned}& \phi=-E d s+E d s \\& \Rightarrow \phi=0 \\& \Rightarrow \phi_{\text {in }}=\phi_{\text {out }}\end{aligned}\) Hence, number of field lines entering is equal to number of field lines leaving. Correct option is (4).
