NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359500
Consider a solid spherical conducting sphere having radius \(R\) and charge \(Q\). The total energy stored in the form of electric field lines is
1 \(\frac{{{Q^2}}}{{8\pi {\varepsilon _0}R}}\)
2 \(\frac{{{Q^2}}}{{\pi {\varepsilon _0}R}}\)
3 \(\frac{{{Q^2}}}{{{\varepsilon _0}R}}\)
4 \(\frac{{{Q^2}}}{{4\pi {\varepsilon _0}}}\)
Explanation:
As the sphere is conducting the total charge \(Q\) resides on its surface. Method-I: The energy stored in the field lines will be equal to work done to assemble charge \(Q\) bringing from infinity, which is equal to \(\frac{{{Q^2}}}{{8\pi {\varepsilon _0}R}}\) Method-II: Electric field inside the conductor is zero. The field lines will be distributed from the surface of the conductor to infinity. The energy density for electric field lines is \(u = \frac{1}{2}{\varepsilon _0}{E^2}\) (E- is the electric field at that point) The energy present in the thickness of a spherical shell of radius r as shown in the figure below is \(dU = udV\) (\(d\) \(V\) is the volume of thin shell) \(dV = 4\pi {r^2}dr\) \(u = \frac{1}{2}{\varepsilon _0}{E^2}\) \(E = \frac{Q}{{4\pi {\varepsilon _0}{r^2}}}\) (From Gauss law) \(\int {dU} = \int\limits_{r = R}^{r = \infty } {\frac{1}{2}{\varepsilon _0}\frac{{{Q^2}}}{{16{\pi ^2}\varepsilon _0^2{r^4}}}\left( {4\pi {r^2}dr} \right)} \) \(U = \frac{{{Q^2}}}{{8\pi {\varepsilon _0}R}}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359501
Charges \(+q\) and \(-q\) are placed at points \(A\) and \(B\) respectively, which are a distance \(2 L\) apart, \(C\) is the mid - point between \(A\) and \(B\). The work done in moving a charge \(+Q\) along the semi circle \(C R D\) is
359502
As per figure a point charged \(+q\) is placed at the origin \(O\) work done in taking another point charge \(-Q\) from the point \(M\) coordinates (0, \(a\)) to another point \(N\) coordinates (\(a\),0) along the straight path \(MN\) is
359503
The amount of work done by electric field in displacing the charges \({q}\) from \({A}\) to \({B}\) in the given figure is ( \({Q=2 \mu C}\) and \({q=1 \mu C}\) )
1 \({-2.4 m J}\)
2 \({6 m J}\)
3 \({3.4 m J}\)
4 \({2.4 m J}\)
Explanation:
\({W=\dfrac{-1}{4 \pi \varepsilon_{0}} q Q\left[\dfrac{1}{5}-\dfrac{1}{3}\right]}\) \({=-9 \times 10^{9} \times 1 \times 10^{-6} \times 2 \times 10^{-6} \times \dfrac{-2}{15}}\) \({=\dfrac{12}{5} \times 10^{-3} {~J}=2.4 {~mJ}}\) Therefore work done by Electric field is \(2.4\,mJ\).
359500
Consider a solid spherical conducting sphere having radius \(R\) and charge \(Q\). The total energy stored in the form of electric field lines is
1 \(\frac{{{Q^2}}}{{8\pi {\varepsilon _0}R}}\)
2 \(\frac{{{Q^2}}}{{\pi {\varepsilon _0}R}}\)
3 \(\frac{{{Q^2}}}{{{\varepsilon _0}R}}\)
4 \(\frac{{{Q^2}}}{{4\pi {\varepsilon _0}}}\)
Explanation:
As the sphere is conducting the total charge \(Q\) resides on its surface. Method-I: The energy stored in the field lines will be equal to work done to assemble charge \(Q\) bringing from infinity, which is equal to \(\frac{{{Q^2}}}{{8\pi {\varepsilon _0}R}}\) Method-II: Electric field inside the conductor is zero. The field lines will be distributed from the surface of the conductor to infinity. The energy density for electric field lines is \(u = \frac{1}{2}{\varepsilon _0}{E^2}\) (E- is the electric field at that point) The energy present in the thickness of a spherical shell of radius r as shown in the figure below is \(dU = udV\) (\(d\) \(V\) is the volume of thin shell) \(dV = 4\pi {r^2}dr\) \(u = \frac{1}{2}{\varepsilon _0}{E^2}\) \(E = \frac{Q}{{4\pi {\varepsilon _0}{r^2}}}\) (From Gauss law) \(\int {dU} = \int\limits_{r = R}^{r = \infty } {\frac{1}{2}{\varepsilon _0}\frac{{{Q^2}}}{{16{\pi ^2}\varepsilon _0^2{r^4}}}\left( {4\pi {r^2}dr} \right)} \) \(U = \frac{{{Q^2}}}{{8\pi {\varepsilon _0}R}}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359501
Charges \(+q\) and \(-q\) are placed at points \(A\) and \(B\) respectively, which are a distance \(2 L\) apart, \(C\) is the mid - point between \(A\) and \(B\). The work done in moving a charge \(+Q\) along the semi circle \(C R D\) is
359502
As per figure a point charged \(+q\) is placed at the origin \(O\) work done in taking another point charge \(-Q\) from the point \(M\) coordinates (0, \(a\)) to another point \(N\) coordinates (\(a\),0) along the straight path \(MN\) is
359503
The amount of work done by electric field in displacing the charges \({q}\) from \({A}\) to \({B}\) in the given figure is ( \({Q=2 \mu C}\) and \({q=1 \mu C}\) )
1 \({-2.4 m J}\)
2 \({6 m J}\)
3 \({3.4 m J}\)
4 \({2.4 m J}\)
Explanation:
\({W=\dfrac{-1}{4 \pi \varepsilon_{0}} q Q\left[\dfrac{1}{5}-\dfrac{1}{3}\right]}\) \({=-9 \times 10^{9} \times 1 \times 10^{-6} \times 2 \times 10^{-6} \times \dfrac{-2}{15}}\) \({=\dfrac{12}{5} \times 10^{-3} {~J}=2.4 {~mJ}}\) Therefore work done by Electric field is \(2.4\,mJ\).
359500
Consider a solid spherical conducting sphere having radius \(R\) and charge \(Q\). The total energy stored in the form of electric field lines is
1 \(\frac{{{Q^2}}}{{8\pi {\varepsilon _0}R}}\)
2 \(\frac{{{Q^2}}}{{\pi {\varepsilon _0}R}}\)
3 \(\frac{{{Q^2}}}{{{\varepsilon _0}R}}\)
4 \(\frac{{{Q^2}}}{{4\pi {\varepsilon _0}}}\)
Explanation:
As the sphere is conducting the total charge \(Q\) resides on its surface. Method-I: The energy stored in the field lines will be equal to work done to assemble charge \(Q\) bringing from infinity, which is equal to \(\frac{{{Q^2}}}{{8\pi {\varepsilon _0}R}}\) Method-II: Electric field inside the conductor is zero. The field lines will be distributed from the surface of the conductor to infinity. The energy density for electric field lines is \(u = \frac{1}{2}{\varepsilon _0}{E^2}\) (E- is the electric field at that point) The energy present in the thickness of a spherical shell of radius r as shown in the figure below is \(dU = udV\) (\(d\) \(V\) is the volume of thin shell) \(dV = 4\pi {r^2}dr\) \(u = \frac{1}{2}{\varepsilon _0}{E^2}\) \(E = \frac{Q}{{4\pi {\varepsilon _0}{r^2}}}\) (From Gauss law) \(\int {dU} = \int\limits_{r = R}^{r = \infty } {\frac{1}{2}{\varepsilon _0}\frac{{{Q^2}}}{{16{\pi ^2}\varepsilon _0^2{r^4}}}\left( {4\pi {r^2}dr} \right)} \) \(U = \frac{{{Q^2}}}{{8\pi {\varepsilon _0}R}}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359501
Charges \(+q\) and \(-q\) are placed at points \(A\) and \(B\) respectively, which are a distance \(2 L\) apart, \(C\) is the mid - point between \(A\) and \(B\). The work done in moving a charge \(+Q\) along the semi circle \(C R D\) is
359502
As per figure a point charged \(+q\) is placed at the origin \(O\) work done in taking another point charge \(-Q\) from the point \(M\) coordinates (0, \(a\)) to another point \(N\) coordinates (\(a\),0) along the straight path \(MN\) is
359503
The amount of work done by electric field in displacing the charges \({q}\) from \({A}\) to \({B}\) in the given figure is ( \({Q=2 \mu C}\) and \({q=1 \mu C}\) )
1 \({-2.4 m J}\)
2 \({6 m J}\)
3 \({3.4 m J}\)
4 \({2.4 m J}\)
Explanation:
\({W=\dfrac{-1}{4 \pi \varepsilon_{0}} q Q\left[\dfrac{1}{5}-\dfrac{1}{3}\right]}\) \({=-9 \times 10^{9} \times 1 \times 10^{-6} \times 2 \times 10^{-6} \times \dfrac{-2}{15}}\) \({=\dfrac{12}{5} \times 10^{-3} {~J}=2.4 {~mJ}}\) Therefore work done by Electric field is \(2.4\,mJ\).
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359500
Consider a solid spherical conducting sphere having radius \(R\) and charge \(Q\). The total energy stored in the form of electric field lines is
1 \(\frac{{{Q^2}}}{{8\pi {\varepsilon _0}R}}\)
2 \(\frac{{{Q^2}}}{{\pi {\varepsilon _0}R}}\)
3 \(\frac{{{Q^2}}}{{{\varepsilon _0}R}}\)
4 \(\frac{{{Q^2}}}{{4\pi {\varepsilon _0}}}\)
Explanation:
As the sphere is conducting the total charge \(Q\) resides on its surface. Method-I: The energy stored in the field lines will be equal to work done to assemble charge \(Q\) bringing from infinity, which is equal to \(\frac{{{Q^2}}}{{8\pi {\varepsilon _0}R}}\) Method-II: Electric field inside the conductor is zero. The field lines will be distributed from the surface of the conductor to infinity. The energy density for electric field lines is \(u = \frac{1}{2}{\varepsilon _0}{E^2}\) (E- is the electric field at that point) The energy present in the thickness of a spherical shell of radius r as shown in the figure below is \(dU = udV\) (\(d\) \(V\) is the volume of thin shell) \(dV = 4\pi {r^2}dr\) \(u = \frac{1}{2}{\varepsilon _0}{E^2}\) \(E = \frac{Q}{{4\pi {\varepsilon _0}{r^2}}}\) (From Gauss law) \(\int {dU} = \int\limits_{r = R}^{r = \infty } {\frac{1}{2}{\varepsilon _0}\frac{{{Q^2}}}{{16{\pi ^2}\varepsilon _0^2{r^4}}}\left( {4\pi {r^2}dr} \right)} \) \(U = \frac{{{Q^2}}}{{8\pi {\varepsilon _0}R}}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359501
Charges \(+q\) and \(-q\) are placed at points \(A\) and \(B\) respectively, which are a distance \(2 L\) apart, \(C\) is the mid - point between \(A\) and \(B\). The work done in moving a charge \(+Q\) along the semi circle \(C R D\) is
359502
As per figure a point charged \(+q\) is placed at the origin \(O\) work done in taking another point charge \(-Q\) from the point \(M\) coordinates (0, \(a\)) to another point \(N\) coordinates (\(a\),0) along the straight path \(MN\) is
359503
The amount of work done by electric field in displacing the charges \({q}\) from \({A}\) to \({B}\) in the given figure is ( \({Q=2 \mu C}\) and \({q=1 \mu C}\) )
1 \({-2.4 m J}\)
2 \({6 m J}\)
3 \({3.4 m J}\)
4 \({2.4 m J}\)
Explanation:
\({W=\dfrac{-1}{4 \pi \varepsilon_{0}} q Q\left[\dfrac{1}{5}-\dfrac{1}{3}\right]}\) \({=-9 \times 10^{9} \times 1 \times 10^{-6} \times 2 \times 10^{-6} \times \dfrac{-2}{15}}\) \({=\dfrac{12}{5} \times 10^{-3} {~J}=2.4 {~mJ}}\) Therefore work done by Electric field is \(2.4\,mJ\).
