358180
The magnitude of point charge due to which the electric field \(30\,cm\) away has the magnitude \(2\,N/C\) will be
1 \(2 \times {10^{ - 11}}C\)
2 \(3 \times {10^{ - 11}}C\)
3 \(5 \times {10^{ - 11}}C\)
4 \(9 \times {10^{ - 11}}C\)
Explanation:
The magnitude of electric field due to a point charge q at a distance r from the charge is \(E = \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{{{r^2}}}\) Here \(E = 2\,N{C^{ - 1}},r = 30\,cm = 0.3\,m,\) \(\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}N{m^2}{C^{ - 2}},q = ?\) \(\therefore 2 = \frac{{(9 \times {{10}^9})q}}{{{{(0.3{\mkern 1mu} )}^2}}}\) \( \Rightarrow {\mkern 1mu} q = \frac{{(2)(0.09)}}{{(9 \times {{10}^9})}}\) \( \Rightarrow {\mkern 1mu} q = \frac{{2 \times 9}}{{9 \times {{10}^9} \times 100}} = 2 \times {10^{ - 11}}C\) Thus, the magnitude of point charge will be \(2 \times {10^{ - 11}}C\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358181
A point charge of \(100\,\mu C\) is placed at \(3\widehat i + 4\widehat j\,m\). Find the electric field intensity due to this charge at a point located at \(9\widehat i + 12\widehat j\,m\).
1 \(8000\,V{m^{ - 1}}\)
2 \(9000\,V{m^{ - 1}}\)
3 \(2250\,V{m^{ - 1}}\)
4 \(4500\,V{m^{ - 1}}\)
Explanation:
\(E = \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{{{{\left| r \right|}^2}}}\) \(\overrightarrow r = (9 - 3)\widehat i + (12 - 4)\widehat j = 6\widehat i + 8\widehat j\) \( \Rightarrow r = \sqrt {{6^2} + {8^2}} = 10m\) \(E = \frac{{9 \times {{10}^9} \times 100 \times {{10}^{ - 6}}}}{{{{10}^2}}} = 9000\,V{m^{ - 1}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358182
The electric field strength at a distance \(r\) from a charge \(q\) is \(E\). What will be electric field strength if the distance of the observation point is increased to \(2r\)?
1 \(E/2\)
2 \(E/4\)
3 \(E/6\)
4 None of the above
Explanation:
Conceptual Question
PHXII01:ELECTRIC CHARGES AND FIELDS
358183
A point charge \({64 \mu C}\) is placed at \({(3 \hat{i}+4 \hat{j}) m}\). Find \({|\vec{E}|}\) at point \({(5 \hat{i}+12 \hat{j}) m}\).
358180
The magnitude of point charge due to which the electric field \(30\,cm\) away has the magnitude \(2\,N/C\) will be
1 \(2 \times {10^{ - 11}}C\)
2 \(3 \times {10^{ - 11}}C\)
3 \(5 \times {10^{ - 11}}C\)
4 \(9 \times {10^{ - 11}}C\)
Explanation:
The magnitude of electric field due to a point charge q at a distance r from the charge is \(E = \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{{{r^2}}}\) Here \(E = 2\,N{C^{ - 1}},r = 30\,cm = 0.3\,m,\) \(\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}N{m^2}{C^{ - 2}},q = ?\) \(\therefore 2 = \frac{{(9 \times {{10}^9})q}}{{{{(0.3{\mkern 1mu} )}^2}}}\) \( \Rightarrow {\mkern 1mu} q = \frac{{(2)(0.09)}}{{(9 \times {{10}^9})}}\) \( \Rightarrow {\mkern 1mu} q = \frac{{2 \times 9}}{{9 \times {{10}^9} \times 100}} = 2 \times {10^{ - 11}}C\) Thus, the magnitude of point charge will be \(2 \times {10^{ - 11}}C\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358181
A point charge of \(100\,\mu C\) is placed at \(3\widehat i + 4\widehat j\,m\). Find the electric field intensity due to this charge at a point located at \(9\widehat i + 12\widehat j\,m\).
1 \(8000\,V{m^{ - 1}}\)
2 \(9000\,V{m^{ - 1}}\)
3 \(2250\,V{m^{ - 1}}\)
4 \(4500\,V{m^{ - 1}}\)
Explanation:
\(E = \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{{{{\left| r \right|}^2}}}\) \(\overrightarrow r = (9 - 3)\widehat i + (12 - 4)\widehat j = 6\widehat i + 8\widehat j\) \( \Rightarrow r = \sqrt {{6^2} + {8^2}} = 10m\) \(E = \frac{{9 \times {{10}^9} \times 100 \times {{10}^{ - 6}}}}{{{{10}^2}}} = 9000\,V{m^{ - 1}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358182
The electric field strength at a distance \(r\) from a charge \(q\) is \(E\). What will be electric field strength if the distance of the observation point is increased to \(2r\)?
1 \(E/2\)
2 \(E/4\)
3 \(E/6\)
4 None of the above
Explanation:
Conceptual Question
PHXII01:ELECTRIC CHARGES AND FIELDS
358183
A point charge \({64 \mu C}\) is placed at \({(3 \hat{i}+4 \hat{j}) m}\). Find \({|\vec{E}|}\) at point \({(5 \hat{i}+12 \hat{j}) m}\).
358180
The magnitude of point charge due to which the electric field \(30\,cm\) away has the magnitude \(2\,N/C\) will be
1 \(2 \times {10^{ - 11}}C\)
2 \(3 \times {10^{ - 11}}C\)
3 \(5 \times {10^{ - 11}}C\)
4 \(9 \times {10^{ - 11}}C\)
Explanation:
The magnitude of electric field due to a point charge q at a distance r from the charge is \(E = \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{{{r^2}}}\) Here \(E = 2\,N{C^{ - 1}},r = 30\,cm = 0.3\,m,\) \(\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}N{m^2}{C^{ - 2}},q = ?\) \(\therefore 2 = \frac{{(9 \times {{10}^9})q}}{{{{(0.3{\mkern 1mu} )}^2}}}\) \( \Rightarrow {\mkern 1mu} q = \frac{{(2)(0.09)}}{{(9 \times {{10}^9})}}\) \( \Rightarrow {\mkern 1mu} q = \frac{{2 \times 9}}{{9 \times {{10}^9} \times 100}} = 2 \times {10^{ - 11}}C\) Thus, the magnitude of point charge will be \(2 \times {10^{ - 11}}C\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358181
A point charge of \(100\,\mu C\) is placed at \(3\widehat i + 4\widehat j\,m\). Find the electric field intensity due to this charge at a point located at \(9\widehat i + 12\widehat j\,m\).
1 \(8000\,V{m^{ - 1}}\)
2 \(9000\,V{m^{ - 1}}\)
3 \(2250\,V{m^{ - 1}}\)
4 \(4500\,V{m^{ - 1}}\)
Explanation:
\(E = \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{{{{\left| r \right|}^2}}}\) \(\overrightarrow r = (9 - 3)\widehat i + (12 - 4)\widehat j = 6\widehat i + 8\widehat j\) \( \Rightarrow r = \sqrt {{6^2} + {8^2}} = 10m\) \(E = \frac{{9 \times {{10}^9} \times 100 \times {{10}^{ - 6}}}}{{{{10}^2}}} = 9000\,V{m^{ - 1}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358182
The electric field strength at a distance \(r\) from a charge \(q\) is \(E\). What will be electric field strength if the distance of the observation point is increased to \(2r\)?
1 \(E/2\)
2 \(E/4\)
3 \(E/6\)
4 None of the above
Explanation:
Conceptual Question
PHXII01:ELECTRIC CHARGES AND FIELDS
358183
A point charge \({64 \mu C}\) is placed at \({(3 \hat{i}+4 \hat{j}) m}\). Find \({|\vec{E}|}\) at point \({(5 \hat{i}+12 \hat{j}) m}\).
358180
The magnitude of point charge due to which the electric field \(30\,cm\) away has the magnitude \(2\,N/C\) will be
1 \(2 \times {10^{ - 11}}C\)
2 \(3 \times {10^{ - 11}}C\)
3 \(5 \times {10^{ - 11}}C\)
4 \(9 \times {10^{ - 11}}C\)
Explanation:
The magnitude of electric field due to a point charge q at a distance r from the charge is \(E = \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{{{r^2}}}\) Here \(E = 2\,N{C^{ - 1}},r = 30\,cm = 0.3\,m,\) \(\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}N{m^2}{C^{ - 2}},q = ?\) \(\therefore 2 = \frac{{(9 \times {{10}^9})q}}{{{{(0.3{\mkern 1mu} )}^2}}}\) \( \Rightarrow {\mkern 1mu} q = \frac{{(2)(0.09)}}{{(9 \times {{10}^9})}}\) \( \Rightarrow {\mkern 1mu} q = \frac{{2 \times 9}}{{9 \times {{10}^9} \times 100}} = 2 \times {10^{ - 11}}C\) Thus, the magnitude of point charge will be \(2 \times {10^{ - 11}}C\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358181
A point charge of \(100\,\mu C\) is placed at \(3\widehat i + 4\widehat j\,m\). Find the electric field intensity due to this charge at a point located at \(9\widehat i + 12\widehat j\,m\).
1 \(8000\,V{m^{ - 1}}\)
2 \(9000\,V{m^{ - 1}}\)
3 \(2250\,V{m^{ - 1}}\)
4 \(4500\,V{m^{ - 1}}\)
Explanation:
\(E = \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{{{{\left| r \right|}^2}}}\) \(\overrightarrow r = (9 - 3)\widehat i + (12 - 4)\widehat j = 6\widehat i + 8\widehat j\) \( \Rightarrow r = \sqrt {{6^2} + {8^2}} = 10m\) \(E = \frac{{9 \times {{10}^9} \times 100 \times {{10}^{ - 6}}}}{{{{10}^2}}} = 9000\,V{m^{ - 1}}\)
PHXII01:ELECTRIC CHARGES AND FIELDS
358182
The electric field strength at a distance \(r\) from a charge \(q\) is \(E\). What will be electric field strength if the distance of the observation point is increased to \(2r\)?
1 \(E/2\)
2 \(E/4\)
3 \(E/6\)
4 None of the above
Explanation:
Conceptual Question
PHXII01:ELECTRIC CHARGES AND FIELDS
358183
A point charge \({64 \mu C}\) is placed at \({(3 \hat{i}+4 \hat{j}) m}\). Find \({|\vec{E}|}\) at point \({(5 \hat{i}+12 \hat{j}) m}\).
