358037
A ring of charge with radius 0.5 \(m\) has \(0.002\,\pi m\) gap. If the ring carries a charge of \( + 1C\), the electric field at the center is
1 \(7.5 \times {10^7}N{C^{ - 1}}\)
2 \(7.2 \times {10^7}N{C^{ - 1}}\)
3 \(6.2 \times {10^7}N{C^{ - 1}}\)
4 \(6.5 \times {10^7}N{C^{ - 1}}\)
Explanation:
As the filed at the centre is zero for a uniformly closed ring, so the field at the centre will be equal and opposite to the field produced by the small element. \(dq = \frac{Q}{{2\pi r}}(0.002\pi )\) \( = \frac{1}{{2\pi (0.5)}} \times \frac{{2\pi }}{{1000}} = 2 \times {10^{ - 3}}{\rm{C}}\) \(E = \frac{{9 \times {{10}^9} \times 2 \times {{10}^{ - 3}}}}{{{{(0.5)}^2}}} = 7.2 \times {10^7}{\rm{N}}{{\rm{C}}^{{\rm{ - 1}}}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358038
Two parallel infinite line charges with linear charge densities \( + \lambda \,C/m\,{\rm{and}}\, - \lambda \,C/m\) are placed at a distance of 2\(R\) in free space. What is the electric field mid-way between the two line charges?
Electric field due to line charge (1) \({E_1} = \frac{\lambda }{{2\pi {\varepsilon _0}R}}\;N{\rm{/}}C\) Electric field due to line charge (2) \({E_2} = \frac{\lambda }{{2\pi {\varepsilon _0}R}}\;N{\rm{/}}C\) \({\overrightarrow E _{net}} = {\overrightarrow E _1} + {\overrightarrow E _2} = \frac{\lambda }{{2\pi {\varepsilon _0}R}}\hat i + \frac{\lambda }{{2\pi {\varepsilon _0}R}}\hat i\) \( = \frac{\lambda }{{\pi {\varepsilon _0}R}}\hat i\,{\rm{N/C}}\)
NEET - 2019
PHXII01:ELECTRIC CHARGES AND FIELDS
358039
A particle of charge \(-q\) and mass \(m\) moves in a circle of radius \(r\) around an infinitely long line charge of linear density \(+\lambda\). Then time period will be given as (Consider \(k\) as Coulomb's constant)
1 \(T=2 \pi r \sqrt{\dfrac{m}{2 k \lambda q}}\)
2 \(T=\dfrac{1}{2 \pi} \sqrt{\dfrac{2 k \lambda q}{m}}\)
3 \(T^{2}=\dfrac{4 \pi^{2} m}{2 k \lambda q} r^{3}\)
4 \(T=\dfrac{1}{2 \pi r} \sqrt{\dfrac{m}{2 k \lambda q}}\)
Explanation:
For a particle to move in a circle, centripetal force is required, this force is equal to the electric force due to the line charge, \(F=m \omega^{2} r=\dfrac{\lambda q}{2 \pi r \varepsilon_{0}} ; \omega=\sqrt{\dfrac{\lambda q}{2 \pi m r^{2} \varepsilon_{0}}}\) \(\left(\because E=\dfrac{1}{2 \pi \varepsilon_{0}} \dfrac{\lambda}{r}\right)\) \( \Rightarrow \frac{{2\pi }}{T} = \sqrt {\frac{{\lambda q}}{{2\pi m{r^2}{\varepsilon _0}}}} \quad \left[ {{\rm{As}},{\rm{ }}\omega = \frac{{2\pi }}{T}} \right]\) \(\Rightarrow \quad T=2 \pi \sqrt{\dfrac{2 \pi m r^{2} \varepsilon_{0}}{\lambda q}}\) or, \(T=2 \pi \sqrt{\dfrac{m r^{2}}{2 \lambda q k}} \quad\left[\because k=\dfrac{1}{4 \pi \varepsilon_{0}}\right]\) \(\therefore \quad T=2 \pi r \sqrt{\dfrac{m}{2 k \lambda q}}\)
JEE - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358040
An infinitely long thin straight wire has uniform charge density of \(\frac{1}{4} \times {10^{ - 2}}C{m^{ - 1}}\). What is the magnitude of electric field at a distance 20 \(cm\) from the axis of the wire ?
1 \(2.25 \times {10^8}N{C^{ - 1}}\)
2 \(9 \times {10^8}N{C^{ - 1}}\)
3 \(1.12 \times {10^8}N{C^{ - 1}}\)
4 \(4.5 \times {10^8}N{C^{ - 1}}\)
Explanation:
Given \(\lambda = \frac{1}{4} \times {10^{ - 2}}C{m^{ - 1}}\) \(x = 20\,cm = 20 \times {10^{ - 2}}m\) From Gauss law, electric field due to an infinitely long wire, \(E = \frac{1}{{2\pi {\varepsilon _0}}}\frac{\lambda }{x} = \frac{2}{{4\pi {\varepsilon _0}}} \cdot \frac{{\rm{\lambda }}}{{\rm{x}}}\) \(or\,E = 2 \times 9 \times {10^9} \times \frac{1}{4} \times {10^{ - 2}} \times \frac{1}{{20 \times {{10}^{ - 2}}}}\) \(or\, E = 0.225 \times {10^9}\) \(\,E = 2.25 \times {10^8}N{C^{ - 1}}\)
358037
A ring of charge with radius 0.5 \(m\) has \(0.002\,\pi m\) gap. If the ring carries a charge of \( + 1C\), the electric field at the center is
1 \(7.5 \times {10^7}N{C^{ - 1}}\)
2 \(7.2 \times {10^7}N{C^{ - 1}}\)
3 \(6.2 \times {10^7}N{C^{ - 1}}\)
4 \(6.5 \times {10^7}N{C^{ - 1}}\)
Explanation:
As the filed at the centre is zero for a uniformly closed ring, so the field at the centre will be equal and opposite to the field produced by the small element. \(dq = \frac{Q}{{2\pi r}}(0.002\pi )\) \( = \frac{1}{{2\pi (0.5)}} \times \frac{{2\pi }}{{1000}} = 2 \times {10^{ - 3}}{\rm{C}}\) \(E = \frac{{9 \times {{10}^9} \times 2 \times {{10}^{ - 3}}}}{{{{(0.5)}^2}}} = 7.2 \times {10^7}{\rm{N}}{{\rm{C}}^{{\rm{ - 1}}}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358038
Two parallel infinite line charges with linear charge densities \( + \lambda \,C/m\,{\rm{and}}\, - \lambda \,C/m\) are placed at a distance of 2\(R\) in free space. What is the electric field mid-way between the two line charges?
Electric field due to line charge (1) \({E_1} = \frac{\lambda }{{2\pi {\varepsilon _0}R}}\;N{\rm{/}}C\) Electric field due to line charge (2) \({E_2} = \frac{\lambda }{{2\pi {\varepsilon _0}R}}\;N{\rm{/}}C\) \({\overrightarrow E _{net}} = {\overrightarrow E _1} + {\overrightarrow E _2} = \frac{\lambda }{{2\pi {\varepsilon _0}R}}\hat i + \frac{\lambda }{{2\pi {\varepsilon _0}R}}\hat i\) \( = \frac{\lambda }{{\pi {\varepsilon _0}R}}\hat i\,{\rm{N/C}}\)
NEET - 2019
PHXII01:ELECTRIC CHARGES AND FIELDS
358039
A particle of charge \(-q\) and mass \(m\) moves in a circle of radius \(r\) around an infinitely long line charge of linear density \(+\lambda\). Then time period will be given as (Consider \(k\) as Coulomb's constant)
1 \(T=2 \pi r \sqrt{\dfrac{m}{2 k \lambda q}}\)
2 \(T=\dfrac{1}{2 \pi} \sqrt{\dfrac{2 k \lambda q}{m}}\)
3 \(T^{2}=\dfrac{4 \pi^{2} m}{2 k \lambda q} r^{3}\)
4 \(T=\dfrac{1}{2 \pi r} \sqrt{\dfrac{m}{2 k \lambda q}}\)
Explanation:
For a particle to move in a circle, centripetal force is required, this force is equal to the electric force due to the line charge, \(F=m \omega^{2} r=\dfrac{\lambda q}{2 \pi r \varepsilon_{0}} ; \omega=\sqrt{\dfrac{\lambda q}{2 \pi m r^{2} \varepsilon_{0}}}\) \(\left(\because E=\dfrac{1}{2 \pi \varepsilon_{0}} \dfrac{\lambda}{r}\right)\) \( \Rightarrow \frac{{2\pi }}{T} = \sqrt {\frac{{\lambda q}}{{2\pi m{r^2}{\varepsilon _0}}}} \quad \left[ {{\rm{As}},{\rm{ }}\omega = \frac{{2\pi }}{T}} \right]\) \(\Rightarrow \quad T=2 \pi \sqrt{\dfrac{2 \pi m r^{2} \varepsilon_{0}}{\lambda q}}\) or, \(T=2 \pi \sqrt{\dfrac{m r^{2}}{2 \lambda q k}} \quad\left[\because k=\dfrac{1}{4 \pi \varepsilon_{0}}\right]\) \(\therefore \quad T=2 \pi r \sqrt{\dfrac{m}{2 k \lambda q}}\)
JEE - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358040
An infinitely long thin straight wire has uniform charge density of \(\frac{1}{4} \times {10^{ - 2}}C{m^{ - 1}}\). What is the magnitude of electric field at a distance 20 \(cm\) from the axis of the wire ?
1 \(2.25 \times {10^8}N{C^{ - 1}}\)
2 \(9 \times {10^8}N{C^{ - 1}}\)
3 \(1.12 \times {10^8}N{C^{ - 1}}\)
4 \(4.5 \times {10^8}N{C^{ - 1}}\)
Explanation:
Given \(\lambda = \frac{1}{4} \times {10^{ - 2}}C{m^{ - 1}}\) \(x = 20\,cm = 20 \times {10^{ - 2}}m\) From Gauss law, electric field due to an infinitely long wire, \(E = \frac{1}{{2\pi {\varepsilon _0}}}\frac{\lambda }{x} = \frac{2}{{4\pi {\varepsilon _0}}} \cdot \frac{{\rm{\lambda }}}{{\rm{x}}}\) \(or\,E = 2 \times 9 \times {10^9} \times \frac{1}{4} \times {10^{ - 2}} \times \frac{1}{{20 \times {{10}^{ - 2}}}}\) \(or\, E = 0.225 \times {10^9}\) \(\,E = 2.25 \times {10^8}N{C^{ - 1}}\)
358037
A ring of charge with radius 0.5 \(m\) has \(0.002\,\pi m\) gap. If the ring carries a charge of \( + 1C\), the electric field at the center is
1 \(7.5 \times {10^7}N{C^{ - 1}}\)
2 \(7.2 \times {10^7}N{C^{ - 1}}\)
3 \(6.2 \times {10^7}N{C^{ - 1}}\)
4 \(6.5 \times {10^7}N{C^{ - 1}}\)
Explanation:
As the filed at the centre is zero for a uniformly closed ring, so the field at the centre will be equal and opposite to the field produced by the small element. \(dq = \frac{Q}{{2\pi r}}(0.002\pi )\) \( = \frac{1}{{2\pi (0.5)}} \times \frac{{2\pi }}{{1000}} = 2 \times {10^{ - 3}}{\rm{C}}\) \(E = \frac{{9 \times {{10}^9} \times 2 \times {{10}^{ - 3}}}}{{{{(0.5)}^2}}} = 7.2 \times {10^7}{\rm{N}}{{\rm{C}}^{{\rm{ - 1}}}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358038
Two parallel infinite line charges with linear charge densities \( + \lambda \,C/m\,{\rm{and}}\, - \lambda \,C/m\) are placed at a distance of 2\(R\) in free space. What is the electric field mid-way between the two line charges?
Electric field due to line charge (1) \({E_1} = \frac{\lambda }{{2\pi {\varepsilon _0}R}}\;N{\rm{/}}C\) Electric field due to line charge (2) \({E_2} = \frac{\lambda }{{2\pi {\varepsilon _0}R}}\;N{\rm{/}}C\) \({\overrightarrow E _{net}} = {\overrightarrow E _1} + {\overrightarrow E _2} = \frac{\lambda }{{2\pi {\varepsilon _0}R}}\hat i + \frac{\lambda }{{2\pi {\varepsilon _0}R}}\hat i\) \( = \frac{\lambda }{{\pi {\varepsilon _0}R}}\hat i\,{\rm{N/C}}\)
NEET - 2019
PHXII01:ELECTRIC CHARGES AND FIELDS
358039
A particle of charge \(-q\) and mass \(m\) moves in a circle of radius \(r\) around an infinitely long line charge of linear density \(+\lambda\). Then time period will be given as (Consider \(k\) as Coulomb's constant)
1 \(T=2 \pi r \sqrt{\dfrac{m}{2 k \lambda q}}\)
2 \(T=\dfrac{1}{2 \pi} \sqrt{\dfrac{2 k \lambda q}{m}}\)
3 \(T^{2}=\dfrac{4 \pi^{2} m}{2 k \lambda q} r^{3}\)
4 \(T=\dfrac{1}{2 \pi r} \sqrt{\dfrac{m}{2 k \lambda q}}\)
Explanation:
For a particle to move in a circle, centripetal force is required, this force is equal to the electric force due to the line charge, \(F=m \omega^{2} r=\dfrac{\lambda q}{2 \pi r \varepsilon_{0}} ; \omega=\sqrt{\dfrac{\lambda q}{2 \pi m r^{2} \varepsilon_{0}}}\) \(\left(\because E=\dfrac{1}{2 \pi \varepsilon_{0}} \dfrac{\lambda}{r}\right)\) \( \Rightarrow \frac{{2\pi }}{T} = \sqrt {\frac{{\lambda q}}{{2\pi m{r^2}{\varepsilon _0}}}} \quad \left[ {{\rm{As}},{\rm{ }}\omega = \frac{{2\pi }}{T}} \right]\) \(\Rightarrow \quad T=2 \pi \sqrt{\dfrac{2 \pi m r^{2} \varepsilon_{0}}{\lambda q}}\) or, \(T=2 \pi \sqrt{\dfrac{m r^{2}}{2 \lambda q k}} \quad\left[\because k=\dfrac{1}{4 \pi \varepsilon_{0}}\right]\) \(\therefore \quad T=2 \pi r \sqrt{\dfrac{m}{2 k \lambda q}}\)
JEE - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358040
An infinitely long thin straight wire has uniform charge density of \(\frac{1}{4} \times {10^{ - 2}}C{m^{ - 1}}\). What is the magnitude of electric field at a distance 20 \(cm\) from the axis of the wire ?
1 \(2.25 \times {10^8}N{C^{ - 1}}\)
2 \(9 \times {10^8}N{C^{ - 1}}\)
3 \(1.12 \times {10^8}N{C^{ - 1}}\)
4 \(4.5 \times {10^8}N{C^{ - 1}}\)
Explanation:
Given \(\lambda = \frac{1}{4} \times {10^{ - 2}}C{m^{ - 1}}\) \(x = 20\,cm = 20 \times {10^{ - 2}}m\) From Gauss law, electric field due to an infinitely long wire, \(E = \frac{1}{{2\pi {\varepsilon _0}}}\frac{\lambda }{x} = \frac{2}{{4\pi {\varepsilon _0}}} \cdot \frac{{\rm{\lambda }}}{{\rm{x}}}\) \(or\,E = 2 \times 9 \times {10^9} \times \frac{1}{4} \times {10^{ - 2}} \times \frac{1}{{20 \times {{10}^{ - 2}}}}\) \(or\, E = 0.225 \times {10^9}\) \(\,E = 2.25 \times {10^8}N{C^{ - 1}}\)
358037
A ring of charge with radius 0.5 \(m\) has \(0.002\,\pi m\) gap. If the ring carries a charge of \( + 1C\), the electric field at the center is
1 \(7.5 \times {10^7}N{C^{ - 1}}\)
2 \(7.2 \times {10^7}N{C^{ - 1}}\)
3 \(6.2 \times {10^7}N{C^{ - 1}}\)
4 \(6.5 \times {10^7}N{C^{ - 1}}\)
Explanation:
As the filed at the centre is zero for a uniformly closed ring, so the field at the centre will be equal and opposite to the field produced by the small element. \(dq = \frac{Q}{{2\pi r}}(0.002\pi )\) \( = \frac{1}{{2\pi (0.5)}} \times \frac{{2\pi }}{{1000}} = 2 \times {10^{ - 3}}{\rm{C}}\) \(E = \frac{{9 \times {{10}^9} \times 2 \times {{10}^{ - 3}}}}{{{{(0.5)}^2}}} = 7.2 \times {10^7}{\rm{N}}{{\rm{C}}^{{\rm{ - 1}}}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358038
Two parallel infinite line charges with linear charge densities \( + \lambda \,C/m\,{\rm{and}}\, - \lambda \,C/m\) are placed at a distance of 2\(R\) in free space. What is the electric field mid-way between the two line charges?
Electric field due to line charge (1) \({E_1} = \frac{\lambda }{{2\pi {\varepsilon _0}R}}\;N{\rm{/}}C\) Electric field due to line charge (2) \({E_2} = \frac{\lambda }{{2\pi {\varepsilon _0}R}}\;N{\rm{/}}C\) \({\overrightarrow E _{net}} = {\overrightarrow E _1} + {\overrightarrow E _2} = \frac{\lambda }{{2\pi {\varepsilon _0}R}}\hat i + \frac{\lambda }{{2\pi {\varepsilon _0}R}}\hat i\) \( = \frac{\lambda }{{\pi {\varepsilon _0}R}}\hat i\,{\rm{N/C}}\)
NEET - 2019
PHXII01:ELECTRIC CHARGES AND FIELDS
358039
A particle of charge \(-q\) and mass \(m\) moves in a circle of radius \(r\) around an infinitely long line charge of linear density \(+\lambda\). Then time period will be given as (Consider \(k\) as Coulomb's constant)
1 \(T=2 \pi r \sqrt{\dfrac{m}{2 k \lambda q}}\)
2 \(T=\dfrac{1}{2 \pi} \sqrt{\dfrac{2 k \lambda q}{m}}\)
3 \(T^{2}=\dfrac{4 \pi^{2} m}{2 k \lambda q} r^{3}\)
4 \(T=\dfrac{1}{2 \pi r} \sqrt{\dfrac{m}{2 k \lambda q}}\)
Explanation:
For a particle to move in a circle, centripetal force is required, this force is equal to the electric force due to the line charge, \(F=m \omega^{2} r=\dfrac{\lambda q}{2 \pi r \varepsilon_{0}} ; \omega=\sqrt{\dfrac{\lambda q}{2 \pi m r^{2} \varepsilon_{0}}}\) \(\left(\because E=\dfrac{1}{2 \pi \varepsilon_{0}} \dfrac{\lambda}{r}\right)\) \( \Rightarrow \frac{{2\pi }}{T} = \sqrt {\frac{{\lambda q}}{{2\pi m{r^2}{\varepsilon _0}}}} \quad \left[ {{\rm{As}},{\rm{ }}\omega = \frac{{2\pi }}{T}} \right]\) \(\Rightarrow \quad T=2 \pi \sqrt{\dfrac{2 \pi m r^{2} \varepsilon_{0}}{\lambda q}}\) or, \(T=2 \pi \sqrt{\dfrac{m r^{2}}{2 \lambda q k}} \quad\left[\because k=\dfrac{1}{4 \pi \varepsilon_{0}}\right]\) \(\therefore \quad T=2 \pi r \sqrt{\dfrac{m}{2 k \lambda q}}\)
JEE - 2024
PHXII01:ELECTRIC CHARGES AND FIELDS
358040
An infinitely long thin straight wire has uniform charge density of \(\frac{1}{4} \times {10^{ - 2}}C{m^{ - 1}}\). What is the magnitude of electric field at a distance 20 \(cm\) from the axis of the wire ?
1 \(2.25 \times {10^8}N{C^{ - 1}}\)
2 \(9 \times {10^8}N{C^{ - 1}}\)
3 \(1.12 \times {10^8}N{C^{ - 1}}\)
4 \(4.5 \times {10^8}N{C^{ - 1}}\)
Explanation:
Given \(\lambda = \frac{1}{4} \times {10^{ - 2}}C{m^{ - 1}}\) \(x = 20\,cm = 20 \times {10^{ - 2}}m\) From Gauss law, electric field due to an infinitely long wire, \(E = \frac{1}{{2\pi {\varepsilon _0}}}\frac{\lambda }{x} = \frac{2}{{4\pi {\varepsilon _0}}} \cdot \frac{{\rm{\lambda }}}{{\rm{x}}}\) \(or\,E = 2 \times 9 \times {10^9} \times \frac{1}{4} \times {10^{ - 2}} \times \frac{1}{{20 \times {{10}^{ - 2}}}}\) \(or\, E = 0.225 \times {10^9}\) \(\,E = 2.25 \times {10^8}N{C^{ - 1}}\)
