359474
A charged \( + q\) is placed at the origin \(O\) of \(X\) - \(Y\) axes are shown in the figure. The work done in taking a charge \(Q\) from \(A\) to \(B\) along the straight line \(AB\) is
Potential at point \(A\) is \({V_A} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{a}\) Potential at point \(B\) is \({V_B} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{b}\) Work done taking a charge \(Q\) from \(A\) to \(B\) is \(W = Q({V_B} - {V_A}) = \frac{{Qq}}{{4\pi {\varepsilon _0}}}\left[ {\frac{1}{b} - \frac{1}{{\rm{a}}}} \right] = \frac{{qQ}}{{4\pi {\varepsilon _0}}}\left( {\frac{{{\rm{a}} - b}}{{{\rm{a}}b}}} \right)\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359475
The displacement of a charge \(Q\) in the electric field \(\overrightarrow E = {e_1}\hat i + {e_2}\hat j + {e_3}\hat k\) is \(\overrightarrow s = a\hat i + b\hat j\). The work done is
By using \(W = Q(\overrightarrow E .\overrightarrow S )\) \( \Rightarrow W = Q\,[({e_1}\hat i + {e_2}\hat j + {e_3}\hat k).(a\hat i + b\hat j)] = \,Q\,({e_1}a + {e_2}b)\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359476
When a positive charge is taken from lower potential to a higher potential point, then its potential energy will
1 Become zero
2 Decrease
3 Increases
4 Remain unchanged
Explanation:
Conceptual Question
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359477
A mass of 1 \(kg\) carrying a charge of 2 \(C\) is accelerated through a potential of 1 \(V\). The velocity acquired by it is
1 \(\sqrt 2 m{s^{ - 1}}\)
2 \(2m{s^{ - 1}}\)
3 \(\frac{1}{{\sqrt 2 }}m{s^{ - 1}}\)
4 \(\frac{1}{2}m{s^{ - 1}}\)
Explanation:
Here,\(q = 2C,m = 1\,kg,V = 1V\) Let the velocity acquired by the mass be \(v\). Then \(\frac{1}{2}m{v^2}\; = \;qV\) or \(v = \sqrt {\frac{{2qV}}{m}} \; = \;\sqrt {\frac{{2\left( {2C} \right)\left( {1V} \right)}}{{1kg}}} \; = \;2m{s^{ - 1}}\)
359474
A charged \( + q\) is placed at the origin \(O\) of \(X\) - \(Y\) axes are shown in the figure. The work done in taking a charge \(Q\) from \(A\) to \(B\) along the straight line \(AB\) is
Potential at point \(A\) is \({V_A} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{a}\) Potential at point \(B\) is \({V_B} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{b}\) Work done taking a charge \(Q\) from \(A\) to \(B\) is \(W = Q({V_B} - {V_A}) = \frac{{Qq}}{{4\pi {\varepsilon _0}}}\left[ {\frac{1}{b} - \frac{1}{{\rm{a}}}} \right] = \frac{{qQ}}{{4\pi {\varepsilon _0}}}\left( {\frac{{{\rm{a}} - b}}{{{\rm{a}}b}}} \right)\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359475
The displacement of a charge \(Q\) in the electric field \(\overrightarrow E = {e_1}\hat i + {e_2}\hat j + {e_3}\hat k\) is \(\overrightarrow s = a\hat i + b\hat j\). The work done is
By using \(W = Q(\overrightarrow E .\overrightarrow S )\) \( \Rightarrow W = Q\,[({e_1}\hat i + {e_2}\hat j + {e_3}\hat k).(a\hat i + b\hat j)] = \,Q\,({e_1}a + {e_2}b)\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359476
When a positive charge is taken from lower potential to a higher potential point, then its potential energy will
1 Become zero
2 Decrease
3 Increases
4 Remain unchanged
Explanation:
Conceptual Question
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359477
A mass of 1 \(kg\) carrying a charge of 2 \(C\) is accelerated through a potential of 1 \(V\). The velocity acquired by it is
1 \(\sqrt 2 m{s^{ - 1}}\)
2 \(2m{s^{ - 1}}\)
3 \(\frac{1}{{\sqrt 2 }}m{s^{ - 1}}\)
4 \(\frac{1}{2}m{s^{ - 1}}\)
Explanation:
Here,\(q = 2C,m = 1\,kg,V = 1V\) Let the velocity acquired by the mass be \(v\). Then \(\frac{1}{2}m{v^2}\; = \;qV\) or \(v = \sqrt {\frac{{2qV}}{m}} \; = \;\sqrt {\frac{{2\left( {2C} \right)\left( {1V} \right)}}{{1kg}}} \; = \;2m{s^{ - 1}}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359474
A charged \( + q\) is placed at the origin \(O\) of \(X\) - \(Y\) axes are shown in the figure. The work done in taking a charge \(Q\) from \(A\) to \(B\) along the straight line \(AB\) is
Potential at point \(A\) is \({V_A} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{a}\) Potential at point \(B\) is \({V_B} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{b}\) Work done taking a charge \(Q\) from \(A\) to \(B\) is \(W = Q({V_B} - {V_A}) = \frac{{Qq}}{{4\pi {\varepsilon _0}}}\left[ {\frac{1}{b} - \frac{1}{{\rm{a}}}} \right] = \frac{{qQ}}{{4\pi {\varepsilon _0}}}\left( {\frac{{{\rm{a}} - b}}{{{\rm{a}}b}}} \right)\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359475
The displacement of a charge \(Q\) in the electric field \(\overrightarrow E = {e_1}\hat i + {e_2}\hat j + {e_3}\hat k\) is \(\overrightarrow s = a\hat i + b\hat j\). The work done is
By using \(W = Q(\overrightarrow E .\overrightarrow S )\) \( \Rightarrow W = Q\,[({e_1}\hat i + {e_2}\hat j + {e_3}\hat k).(a\hat i + b\hat j)] = \,Q\,({e_1}a + {e_2}b)\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359476
When a positive charge is taken from lower potential to a higher potential point, then its potential energy will
1 Become zero
2 Decrease
3 Increases
4 Remain unchanged
Explanation:
Conceptual Question
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359477
A mass of 1 \(kg\) carrying a charge of 2 \(C\) is accelerated through a potential of 1 \(V\). The velocity acquired by it is
1 \(\sqrt 2 m{s^{ - 1}}\)
2 \(2m{s^{ - 1}}\)
3 \(\frac{1}{{\sqrt 2 }}m{s^{ - 1}}\)
4 \(\frac{1}{2}m{s^{ - 1}}\)
Explanation:
Here,\(q = 2C,m = 1\,kg,V = 1V\) Let the velocity acquired by the mass be \(v\). Then \(\frac{1}{2}m{v^2}\; = \;qV\) or \(v = \sqrt {\frac{{2qV}}{m}} \; = \;\sqrt {\frac{{2\left( {2C} \right)\left( {1V} \right)}}{{1kg}}} \; = \;2m{s^{ - 1}}\)
359474
A charged \( + q\) is placed at the origin \(O\) of \(X\) - \(Y\) axes are shown in the figure. The work done in taking a charge \(Q\) from \(A\) to \(B\) along the straight line \(AB\) is
Potential at point \(A\) is \({V_A} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{a}\) Potential at point \(B\) is \({V_B} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{b}\) Work done taking a charge \(Q\) from \(A\) to \(B\) is \(W = Q({V_B} - {V_A}) = \frac{{Qq}}{{4\pi {\varepsilon _0}}}\left[ {\frac{1}{b} - \frac{1}{{\rm{a}}}} \right] = \frac{{qQ}}{{4\pi {\varepsilon _0}}}\left( {\frac{{{\rm{a}} - b}}{{{\rm{a}}b}}} \right)\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359475
The displacement of a charge \(Q\) in the electric field \(\overrightarrow E = {e_1}\hat i + {e_2}\hat j + {e_3}\hat k\) is \(\overrightarrow s = a\hat i + b\hat j\). The work done is
By using \(W = Q(\overrightarrow E .\overrightarrow S )\) \( \Rightarrow W = Q\,[({e_1}\hat i + {e_2}\hat j + {e_3}\hat k).(a\hat i + b\hat j)] = \,Q\,({e_1}a + {e_2}b)\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359476
When a positive charge is taken from lower potential to a higher potential point, then its potential energy will
1 Become zero
2 Decrease
3 Increases
4 Remain unchanged
Explanation:
Conceptual Question
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359477
A mass of 1 \(kg\) carrying a charge of 2 \(C\) is accelerated through a potential of 1 \(V\). The velocity acquired by it is
1 \(\sqrt 2 m{s^{ - 1}}\)
2 \(2m{s^{ - 1}}\)
3 \(\frac{1}{{\sqrt 2 }}m{s^{ - 1}}\)
4 \(\frac{1}{2}m{s^{ - 1}}\)
Explanation:
Here,\(q = 2C,m = 1\,kg,V = 1V\) Let the velocity acquired by the mass be \(v\). Then \(\frac{1}{2}m{v^2}\; = \;qV\) or \(v = \sqrt {\frac{{2qV}}{m}} \; = \;\sqrt {\frac{{2\left( {2C} \right)\left( {1V} \right)}}{{1kg}}} \; = \;2m{s^{ - 1}}\)
