NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII01:ELECTRIC CHARGES AND FIELDS
358142
A neutral water molecule \(\left( {{H_2}O} \right)\) in its vapour state has an electric dipole moment of magnitude \(6.4 \times {10^{ - 30}}C - m\). How far apart are the molecules centres of positive and negative charges?
1 \(4\;m\)
2 \(4\;mm\)
3 \(4\mu m\)
4 \(4\,pm\)
Explanation:
There are 10 electrons and 10 protons in a neutral water molecule. So, its dipole moment is \(p=q(2 l)=10 e(2 l)\) Hence length of the dipole \(i.e.,\) distance between centres of positive and negative charges is \(2l = \frac{p}{{10e}} = \frac{{6.4 \times {{10}^{ - 30}}}}{{10 \times 1.6 \times {{10}^{ - 19}}}}\) \( = 4 \times {10^{ - 12}}\;m = 4\,pm\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358143
The electric intensity due to a dipole of length 1\(m\) and having charge of \(100\,\mu C\) at a point on the axial line at a distance 2\(m\) from one of the charges in air, is
1 \(1.25 \times {10^5}\,N{\rm{/}}C\)
2 \(9.28 \times {10^7}\,N{\rm{/}}C\)
3 \(13.1 \times {10^{11}}\,N{\rm{/}}C\)
4 \(20.5 \times {10^7}\,N{\rm{/}}C\)
Explanation:
Given that Length of the dipole \((d) = 1\,m\) Charge on the dipole \(\left( q \right) = 100\,\mu C = {10^{ - 4}}C\) and distance of the point on the axis from the mid-point of the dipole \(\left( r \right) = 2 + 0.5 = 2.5\,\,m.\) We know that the electric field intensity due to dipole on the given point (\(E\)) As \(r\) and \(d\) are comparable to each other. \( = \frac{1}{{4\pi {\varepsilon _0}}} \times \frac{{2\left( {q.d} \right)r}}{{{{\left( {{r^2} - \left( {\frac{d}{2}} \right)} \right)}^2}}}\) \( = 9 \times {10^9} \times \frac{{2\left( {{{10}^{ - 4}} \times 1} \right) \times 2.5}}{{{{\left[ {{{\left( {2.5} \right)}^2} - {{\left( {0.5} \right)}^2}} \right]}^2}}}\) \( = \frac{{18 \times 2.5 \times {{10}^5}}}{{36}} = 1.25 \times {10^5}\,\,N{\rm{/}}C\) (\(k = 1\) for air)
PHXII01:ELECTRIC CHARGES AND FIELDS
358144
Two charges \( \pm 20\,\mu C\) are placed 10 \(mm\) apart. The electric field at point \(P\), On the axis of the dipole 10 \(cm\) away from its centre \(O\) on the side of the positive charge is
358142
A neutral water molecule \(\left( {{H_2}O} \right)\) in its vapour state has an electric dipole moment of magnitude \(6.4 \times {10^{ - 30}}C - m\). How far apart are the molecules centres of positive and negative charges?
1 \(4\;m\)
2 \(4\;mm\)
3 \(4\mu m\)
4 \(4\,pm\)
Explanation:
There are 10 electrons and 10 protons in a neutral water molecule. So, its dipole moment is \(p=q(2 l)=10 e(2 l)\) Hence length of the dipole \(i.e.,\) distance between centres of positive and negative charges is \(2l = \frac{p}{{10e}} = \frac{{6.4 \times {{10}^{ - 30}}}}{{10 \times 1.6 \times {{10}^{ - 19}}}}\) \( = 4 \times {10^{ - 12}}\;m = 4\,pm\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358143
The electric intensity due to a dipole of length 1\(m\) and having charge of \(100\,\mu C\) at a point on the axial line at a distance 2\(m\) from one of the charges in air, is
1 \(1.25 \times {10^5}\,N{\rm{/}}C\)
2 \(9.28 \times {10^7}\,N{\rm{/}}C\)
3 \(13.1 \times {10^{11}}\,N{\rm{/}}C\)
4 \(20.5 \times {10^7}\,N{\rm{/}}C\)
Explanation:
Given that Length of the dipole \((d) = 1\,m\) Charge on the dipole \(\left( q \right) = 100\,\mu C = {10^{ - 4}}C\) and distance of the point on the axis from the mid-point of the dipole \(\left( r \right) = 2 + 0.5 = 2.5\,\,m.\) We know that the electric field intensity due to dipole on the given point (\(E\)) As \(r\) and \(d\) are comparable to each other. \( = \frac{1}{{4\pi {\varepsilon _0}}} \times \frac{{2\left( {q.d} \right)r}}{{{{\left( {{r^2} - \left( {\frac{d}{2}} \right)} \right)}^2}}}\) \( = 9 \times {10^9} \times \frac{{2\left( {{{10}^{ - 4}} \times 1} \right) \times 2.5}}{{{{\left[ {{{\left( {2.5} \right)}^2} - {{\left( {0.5} \right)}^2}} \right]}^2}}}\) \( = \frac{{18 \times 2.5 \times {{10}^5}}}{{36}} = 1.25 \times {10^5}\,\,N{\rm{/}}C\) (\(k = 1\) for air)
PHXII01:ELECTRIC CHARGES AND FIELDS
358144
Two charges \( \pm 20\,\mu C\) are placed 10 \(mm\) apart. The electric field at point \(P\), On the axis of the dipole 10 \(cm\) away from its centre \(O\) on the side of the positive charge is
358142
A neutral water molecule \(\left( {{H_2}O} \right)\) in its vapour state has an electric dipole moment of magnitude \(6.4 \times {10^{ - 30}}C - m\). How far apart are the molecules centres of positive and negative charges?
1 \(4\;m\)
2 \(4\;mm\)
3 \(4\mu m\)
4 \(4\,pm\)
Explanation:
There are 10 electrons and 10 protons in a neutral water molecule. So, its dipole moment is \(p=q(2 l)=10 e(2 l)\) Hence length of the dipole \(i.e.,\) distance between centres of positive and negative charges is \(2l = \frac{p}{{10e}} = \frac{{6.4 \times {{10}^{ - 30}}}}{{10 \times 1.6 \times {{10}^{ - 19}}}}\) \( = 4 \times {10^{ - 12}}\;m = 4\,pm\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358143
The electric intensity due to a dipole of length 1\(m\) and having charge of \(100\,\mu C\) at a point on the axial line at a distance 2\(m\) from one of the charges in air, is
1 \(1.25 \times {10^5}\,N{\rm{/}}C\)
2 \(9.28 \times {10^7}\,N{\rm{/}}C\)
3 \(13.1 \times {10^{11}}\,N{\rm{/}}C\)
4 \(20.5 \times {10^7}\,N{\rm{/}}C\)
Explanation:
Given that Length of the dipole \((d) = 1\,m\) Charge on the dipole \(\left( q \right) = 100\,\mu C = {10^{ - 4}}C\) and distance of the point on the axis from the mid-point of the dipole \(\left( r \right) = 2 + 0.5 = 2.5\,\,m.\) We know that the electric field intensity due to dipole on the given point (\(E\)) As \(r\) and \(d\) are comparable to each other. \( = \frac{1}{{4\pi {\varepsilon _0}}} \times \frac{{2\left( {q.d} \right)r}}{{{{\left( {{r^2} - \left( {\frac{d}{2}} \right)} \right)}^2}}}\) \( = 9 \times {10^9} \times \frac{{2\left( {{{10}^{ - 4}} \times 1} \right) \times 2.5}}{{{{\left[ {{{\left( {2.5} \right)}^2} - {{\left( {0.5} \right)}^2}} \right]}^2}}}\) \( = \frac{{18 \times 2.5 \times {{10}^5}}}{{36}} = 1.25 \times {10^5}\,\,N{\rm{/}}C\) (\(k = 1\) for air)
PHXII01:ELECTRIC CHARGES AND FIELDS
358144
Two charges \( \pm 20\,\mu C\) are placed 10 \(mm\) apart. The electric field at point \(P\), On the axis of the dipole 10 \(cm\) away from its centre \(O\) on the side of the positive charge is
358142
A neutral water molecule \(\left( {{H_2}O} \right)\) in its vapour state has an electric dipole moment of magnitude \(6.4 \times {10^{ - 30}}C - m\). How far apart are the molecules centres of positive and negative charges?
1 \(4\;m\)
2 \(4\;mm\)
3 \(4\mu m\)
4 \(4\,pm\)
Explanation:
There are 10 electrons and 10 protons in a neutral water molecule. So, its dipole moment is \(p=q(2 l)=10 e(2 l)\) Hence length of the dipole \(i.e.,\) distance between centres of positive and negative charges is \(2l = \frac{p}{{10e}} = \frac{{6.4 \times {{10}^{ - 30}}}}{{10 \times 1.6 \times {{10}^{ - 19}}}}\) \( = 4 \times {10^{ - 12}}\;m = 4\,pm\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358143
The electric intensity due to a dipole of length 1\(m\) and having charge of \(100\,\mu C\) at a point on the axial line at a distance 2\(m\) from one of the charges in air, is
1 \(1.25 \times {10^5}\,N{\rm{/}}C\)
2 \(9.28 \times {10^7}\,N{\rm{/}}C\)
3 \(13.1 \times {10^{11}}\,N{\rm{/}}C\)
4 \(20.5 \times {10^7}\,N{\rm{/}}C\)
Explanation:
Given that Length of the dipole \((d) = 1\,m\) Charge on the dipole \(\left( q \right) = 100\,\mu C = {10^{ - 4}}C\) and distance of the point on the axis from the mid-point of the dipole \(\left( r \right) = 2 + 0.5 = 2.5\,\,m.\) We know that the electric field intensity due to dipole on the given point (\(E\)) As \(r\) and \(d\) are comparable to each other. \( = \frac{1}{{4\pi {\varepsilon _0}}} \times \frac{{2\left( {q.d} \right)r}}{{{{\left( {{r^2} - \left( {\frac{d}{2}} \right)} \right)}^2}}}\) \( = 9 \times {10^9} \times \frac{{2\left( {{{10}^{ - 4}} \times 1} \right) \times 2.5}}{{{{\left[ {{{\left( {2.5} \right)}^2} - {{\left( {0.5} \right)}^2}} \right]}^2}}}\) \( = \frac{{18 \times 2.5 \times {{10}^5}}}{{36}} = 1.25 \times {10^5}\,\,N{\rm{/}}C\) (\(k = 1\) for air)
PHXII01:ELECTRIC CHARGES AND FIELDS
358144
Two charges \( \pm 20\,\mu C\) are placed 10 \(mm\) apart. The electric field at point \(P\), On the axis of the dipole 10 \(cm\) away from its centre \(O\) on the side of the positive charge is
