NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXII01:ELECTRIC CHARGES AND FIELDS
358020
A conducting sphere of radius 10 \(cm\) has an unknown charge. If the electric field at a distance 20 \(cm\) from the centre of the sphere is \(1.2 \times {10^3}\,\,N{C^{ - 1}}\) and points radially inwards. The net charge on the sphere is
1 \( - 4.5 \times {10^{ - 9}}C\)
2 \(4.5 \times {10^9}C\)
3 \( - 5.3 \times {10^{ - 9}}C\)
4 \(5.3 \times {10^9}C\)
Explanation:
Here, radius of sphere \( = 10\,cm\) Distance of point from the centre of the sphere \(r = 20{\rm{ }}cm{\rm{ }} = 0.2{\rm{ }}m\) Electric field, \(E = 1.2 \times {10^3}N{C^{ - 1}}\) As \(E = \frac{q}{{4\pi {\varepsilon _0}{r^2}}}\) \(\therefore q = \left( {4\pi {\varepsilon _0}{r^2}} \right)E\) \( = \frac{{{{\left( {0.2} \right)}^2} \times \left( { - 1.2 \times {{10}^3}} \right)}}{{9 \times {{10}^9}}} = - 5.3 \times {10^{ - 9}}C\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358021
A spherical conductor of radius 2 \(cm\) is uniformly charged with 3 \(nC\). What is the electric field at a distance of 3 \(cm\) from the centre of the sphere?
1 \(3 \times {10^4}V{m^{ - 1}}\)
2 \(3 \times {10^6}V{m^{ - 1}}\)
3 \(3 \times {10^{ - 4}}V{m^{ - 1}}\)
4 \(3\,V{m^{ - 1}}\)
Explanation:
Here,\(Q = 3\,n\,C = 3 \times {10^{ - 9}}C\) \(R = 2\,cm = 2 \times {10^{ - 2}}m\) At a point 3 \(cm\) from the centre, i.e.,\(r = 3\,cm = 3 \times {10^{ - 2}}m\) \(\therefore \) Electric field, \(E = \frac{1}{{4\pi {\varepsilon _0}}}\frac{Q}{{{r^2}}} = \frac{{9 \times {{10}^9} \times 3 \times {{10}^{ - 9}}}}{{{{(3 \times {{10}^{ - 2}})}^2}}}\) \( = 3 \times {10^4}V{m^{ - 1}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358022
Assertion : The surface charge densities of two spherical conductors of different radii are equal. Then the electric field intensities near their surface are also equal. Reason : Surface charge density is equal to charge per unit area.
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:
\(\vec{E}=\dfrac{\sigma}{E_{0}}\) Reason just defines \(\sigma\). The electric field intensities are equal near the surface of two spherical conductors with equal surface charge densities. So correct option is (2).
PHXII01:ELECTRIC CHARGES AND FIELDS
358023
The parallel plane sheets 1 and 2 carry uniform charge densities \({\sigma _{1\,}}{\rm{and}}\,{\sigma _2}\) as shown in the figure. The magnitude of the resultant electric field in the region marked III is \(({\sigma _{1\,}} > \,{\sigma _2})\)
The field produced by an infinite sheet of charge density \(\sigma \) is \(\frac{\sigma }{{2{\varepsilon _0}}}\). The resultant electric field is given by \(E = {E_1} + {E_2} \Rightarrow \) \(E = \frac{{{\sigma _1}}}{{2{\varepsilon _0}}} + \frac{{{\sigma _2}}}{{2{\varepsilon _0}}} = \frac{{{\sigma _1} + {\sigma _2}}}{{2{\varepsilon _0}}}\)
358020
A conducting sphere of radius 10 \(cm\) has an unknown charge. If the electric field at a distance 20 \(cm\) from the centre of the sphere is \(1.2 \times {10^3}\,\,N{C^{ - 1}}\) and points radially inwards. The net charge on the sphere is
1 \( - 4.5 \times {10^{ - 9}}C\)
2 \(4.5 \times {10^9}C\)
3 \( - 5.3 \times {10^{ - 9}}C\)
4 \(5.3 \times {10^9}C\)
Explanation:
Here, radius of sphere \( = 10\,cm\) Distance of point from the centre of the sphere \(r = 20{\rm{ }}cm{\rm{ }} = 0.2{\rm{ }}m\) Electric field, \(E = 1.2 \times {10^3}N{C^{ - 1}}\) As \(E = \frac{q}{{4\pi {\varepsilon _0}{r^2}}}\) \(\therefore q = \left( {4\pi {\varepsilon _0}{r^2}} \right)E\) \( = \frac{{{{\left( {0.2} \right)}^2} \times \left( { - 1.2 \times {{10}^3}} \right)}}{{9 \times {{10}^9}}} = - 5.3 \times {10^{ - 9}}C\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358021
A spherical conductor of radius 2 \(cm\) is uniformly charged with 3 \(nC\). What is the electric field at a distance of 3 \(cm\) from the centre of the sphere?
1 \(3 \times {10^4}V{m^{ - 1}}\)
2 \(3 \times {10^6}V{m^{ - 1}}\)
3 \(3 \times {10^{ - 4}}V{m^{ - 1}}\)
4 \(3\,V{m^{ - 1}}\)
Explanation:
Here,\(Q = 3\,n\,C = 3 \times {10^{ - 9}}C\) \(R = 2\,cm = 2 \times {10^{ - 2}}m\) At a point 3 \(cm\) from the centre, i.e.,\(r = 3\,cm = 3 \times {10^{ - 2}}m\) \(\therefore \) Electric field, \(E = \frac{1}{{4\pi {\varepsilon _0}}}\frac{Q}{{{r^2}}} = \frac{{9 \times {{10}^9} \times 3 \times {{10}^{ - 9}}}}{{{{(3 \times {{10}^{ - 2}})}^2}}}\) \( = 3 \times {10^4}V{m^{ - 1}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358022
Assertion : The surface charge densities of two spherical conductors of different radii are equal. Then the electric field intensities near their surface are also equal. Reason : Surface charge density is equal to charge per unit area.
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:
\(\vec{E}=\dfrac{\sigma}{E_{0}}\) Reason just defines \(\sigma\). The electric field intensities are equal near the surface of two spherical conductors with equal surface charge densities. So correct option is (2).
PHXII01:ELECTRIC CHARGES AND FIELDS
358023
The parallel plane sheets 1 and 2 carry uniform charge densities \({\sigma _{1\,}}{\rm{and}}\,{\sigma _2}\) as shown in the figure. The magnitude of the resultant electric field in the region marked III is \(({\sigma _{1\,}} > \,{\sigma _2})\)
The field produced by an infinite sheet of charge density \(\sigma \) is \(\frac{\sigma }{{2{\varepsilon _0}}}\). The resultant electric field is given by \(E = {E_1} + {E_2} \Rightarrow \) \(E = \frac{{{\sigma _1}}}{{2{\varepsilon _0}}} + \frac{{{\sigma _2}}}{{2{\varepsilon _0}}} = \frac{{{\sigma _1} + {\sigma _2}}}{{2{\varepsilon _0}}}\)
358020
A conducting sphere of radius 10 \(cm\) has an unknown charge. If the electric field at a distance 20 \(cm\) from the centre of the sphere is \(1.2 \times {10^3}\,\,N{C^{ - 1}}\) and points radially inwards. The net charge on the sphere is
1 \( - 4.5 \times {10^{ - 9}}C\)
2 \(4.5 \times {10^9}C\)
3 \( - 5.3 \times {10^{ - 9}}C\)
4 \(5.3 \times {10^9}C\)
Explanation:
Here, radius of sphere \( = 10\,cm\) Distance of point from the centre of the sphere \(r = 20{\rm{ }}cm{\rm{ }} = 0.2{\rm{ }}m\) Electric field, \(E = 1.2 \times {10^3}N{C^{ - 1}}\) As \(E = \frac{q}{{4\pi {\varepsilon _0}{r^2}}}\) \(\therefore q = \left( {4\pi {\varepsilon _0}{r^2}} \right)E\) \( = \frac{{{{\left( {0.2} \right)}^2} \times \left( { - 1.2 \times {{10}^3}} \right)}}{{9 \times {{10}^9}}} = - 5.3 \times {10^{ - 9}}C\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358021
A spherical conductor of radius 2 \(cm\) is uniformly charged with 3 \(nC\). What is the electric field at a distance of 3 \(cm\) from the centre of the sphere?
1 \(3 \times {10^4}V{m^{ - 1}}\)
2 \(3 \times {10^6}V{m^{ - 1}}\)
3 \(3 \times {10^{ - 4}}V{m^{ - 1}}\)
4 \(3\,V{m^{ - 1}}\)
Explanation:
Here,\(Q = 3\,n\,C = 3 \times {10^{ - 9}}C\) \(R = 2\,cm = 2 \times {10^{ - 2}}m\) At a point 3 \(cm\) from the centre, i.e.,\(r = 3\,cm = 3 \times {10^{ - 2}}m\) \(\therefore \) Electric field, \(E = \frac{1}{{4\pi {\varepsilon _0}}}\frac{Q}{{{r^2}}} = \frac{{9 \times {{10}^9} \times 3 \times {{10}^{ - 9}}}}{{{{(3 \times {{10}^{ - 2}})}^2}}}\) \( = 3 \times {10^4}V{m^{ - 1}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358022
Assertion : The surface charge densities of two spherical conductors of different radii are equal. Then the electric field intensities near their surface are also equal. Reason : Surface charge density is equal to charge per unit area.
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:
\(\vec{E}=\dfrac{\sigma}{E_{0}}\) Reason just defines \(\sigma\). The electric field intensities are equal near the surface of two spherical conductors with equal surface charge densities. So correct option is (2).
PHXII01:ELECTRIC CHARGES AND FIELDS
358023
The parallel plane sheets 1 and 2 carry uniform charge densities \({\sigma _{1\,}}{\rm{and}}\,{\sigma _2}\) as shown in the figure. The magnitude of the resultant electric field in the region marked III is \(({\sigma _{1\,}} > \,{\sigma _2})\)
The field produced by an infinite sheet of charge density \(\sigma \) is \(\frac{\sigma }{{2{\varepsilon _0}}}\). The resultant electric field is given by \(E = {E_1} + {E_2} \Rightarrow \) \(E = \frac{{{\sigma _1}}}{{2{\varepsilon _0}}} + \frac{{{\sigma _2}}}{{2{\varepsilon _0}}} = \frac{{{\sigma _1} + {\sigma _2}}}{{2{\varepsilon _0}}}\)
358020
A conducting sphere of radius 10 \(cm\) has an unknown charge. If the electric field at a distance 20 \(cm\) from the centre of the sphere is \(1.2 \times {10^3}\,\,N{C^{ - 1}}\) and points radially inwards. The net charge on the sphere is
1 \( - 4.5 \times {10^{ - 9}}C\)
2 \(4.5 \times {10^9}C\)
3 \( - 5.3 \times {10^{ - 9}}C\)
4 \(5.3 \times {10^9}C\)
Explanation:
Here, radius of sphere \( = 10\,cm\) Distance of point from the centre of the sphere \(r = 20{\rm{ }}cm{\rm{ }} = 0.2{\rm{ }}m\) Electric field, \(E = 1.2 \times {10^3}N{C^{ - 1}}\) As \(E = \frac{q}{{4\pi {\varepsilon _0}{r^2}}}\) \(\therefore q = \left( {4\pi {\varepsilon _0}{r^2}} \right)E\) \( = \frac{{{{\left( {0.2} \right)}^2} \times \left( { - 1.2 \times {{10}^3}} \right)}}{{9 \times {{10}^9}}} = - 5.3 \times {10^{ - 9}}C\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358021
A spherical conductor of radius 2 \(cm\) is uniformly charged with 3 \(nC\). What is the electric field at a distance of 3 \(cm\) from the centre of the sphere?
1 \(3 \times {10^4}V{m^{ - 1}}\)
2 \(3 \times {10^6}V{m^{ - 1}}\)
3 \(3 \times {10^{ - 4}}V{m^{ - 1}}\)
4 \(3\,V{m^{ - 1}}\)
Explanation:
Here,\(Q = 3\,n\,C = 3 \times {10^{ - 9}}C\) \(R = 2\,cm = 2 \times {10^{ - 2}}m\) At a point 3 \(cm\) from the centre, i.e.,\(r = 3\,cm = 3 \times {10^{ - 2}}m\) \(\therefore \) Electric field, \(E = \frac{1}{{4\pi {\varepsilon _0}}}\frac{Q}{{{r^2}}} = \frac{{9 \times {{10}^9} \times 3 \times {{10}^{ - 9}}}}{{{{(3 \times {{10}^{ - 2}})}^2}}}\) \( = 3 \times {10^4}V{m^{ - 1}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358022
Assertion : The surface charge densities of two spherical conductors of different radii are equal. Then the electric field intensities near their surface are also equal. Reason : Surface charge density is equal to charge per unit area.
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:
\(\vec{E}=\dfrac{\sigma}{E_{0}}\) Reason just defines \(\sigma\). The electric field intensities are equal near the surface of two spherical conductors with equal surface charge densities. So correct option is (2).
PHXII01:ELECTRIC CHARGES AND FIELDS
358023
The parallel plane sheets 1 and 2 carry uniform charge densities \({\sigma _{1\,}}{\rm{and}}\,{\sigma _2}\) as shown in the figure. The magnitude of the resultant electric field in the region marked III is \(({\sigma _{1\,}} > \,{\sigma _2})\)
The field produced by an infinite sheet of charge density \(\sigma \) is \(\frac{\sigma }{{2{\varepsilon _0}}}\). The resultant electric field is given by \(E = {E_1} + {E_2} \Rightarrow \) \(E = \frac{{{\sigma _1}}}{{2{\varepsilon _0}}} + \frac{{{\sigma _2}}}{{2{\varepsilon _0}}} = \frac{{{\sigma _1} + {\sigma _2}}}{{2{\varepsilon _0}}}\)
