359491
If identical charges \(\left( { - q} \right)\) are placed at each corner of a cube of side \(b,\) then electric potential energy of charge \(\left( { + q} \right)\) which is placed at centre of the cube will be
Length of the longest diagonal of a cube having each side \(b\) is \(\sqrt 3 \,b.\) So distance of centre of cube from each vertex is \(\frac{{\sqrt 3 \,b}}{2}.\) Potential energy of the given system is \({U_{Sys}} = 8 \times \left\{ {\frac{1}{{4\pi {\varepsilon _0}}}.\frac{{\left( { - q} \right)\left( q \right)}}{{\sqrt 3 b/2}}} \right\} = \frac{{ - 4{q^2}}}{{\sqrt 3 \pi {\varepsilon _0}b}}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359492
Consider a conducting spherical shell of radius \(R\). A charge \(Q\) is given onto the surface of the sphere. The total energy of the shell is
1 \(\frac{{{Q^2}}}{{4\pi {\varepsilon _0}}}\)
2 \(\frac{{{Q^2}}}{{4\pi {\varepsilon _0}R}}\)
3 \(\frac{{{Q^2}}}{{8\pi {\varepsilon _0}R}}\)
4 \(\frac{{{Q^2}}}{{\pi {\varepsilon _0}R}}\)
Explanation:
The energy of the shell is equal to the work done by an external agent to bring the charge from \(\infty \) to the surface. Let \(q\) be the charge already formed on the surface. The small work done \(d{w_{ext}}\) to bring \(dq\) from \(\infty \) to the surface is \(d{W_{ele}} = - \left( {dU} \right) = - \left( {{U_f} - {U_i}} \right)\) \( = - \left[ {\frac{{qdq}}{{4\pi {\varepsilon _0}R}} - \frac{{qdq}}{{4\pi {\varepsilon _0}\left( \infty \right)}}} \right]\) \(d{W_{ele}} = \frac{{ - qdq}}{{4\pi {\varepsilon _0}R}}\) \(d{W_{ext}} = - d{W_{ele}} = \frac{{qdq}}{{4\pi {\varepsilon _0}R}}\) \( \Rightarrow {W_{ert}} = \frac{1}{{4\pi {\varepsilon _0}R}}\int_0^Q {qdq} \) \({W_{ext}} = \frac{{{Q^2}}}{{8\pi {\varepsilon _0}R}}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359493
Three charges are placed at the vertex of an equilateral triangle as shown in figure. For what value of \(Q\) the electrostatic potential energy of the system is zero?
1 \(-q\)
2 \(q / 2\)
3 \(-2 q\)
4 \(\dfrac{-q}{2}\)
Explanation:
Electrostatic potential energy of the given system is given as. \(\begin{array}{ll}U=\dfrac{1}{4 \pi \varepsilon_{0}} \times \dfrac{1}{a}\left(Q q+Q q+q^{2}\right) \\2 Q q+q^{2}=0 & 2 Q q=-q^{2} \\2 Q=-q & Q=\dfrac{-q}{2}\end{array}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359494
Three charges \({Q,+q}\) and \({+q}\) are placed at the vertices of a right-angled isosceles triangle as shown. The net electrostatic energy of the configuration is zero if \({Q}\) is equal to
359491
If identical charges \(\left( { - q} \right)\) are placed at each corner of a cube of side \(b,\) then electric potential energy of charge \(\left( { + q} \right)\) which is placed at centre of the cube will be
Length of the longest diagonal of a cube having each side \(b\) is \(\sqrt 3 \,b.\) So distance of centre of cube from each vertex is \(\frac{{\sqrt 3 \,b}}{2}.\) Potential energy of the given system is \({U_{Sys}} = 8 \times \left\{ {\frac{1}{{4\pi {\varepsilon _0}}}.\frac{{\left( { - q} \right)\left( q \right)}}{{\sqrt 3 b/2}}} \right\} = \frac{{ - 4{q^2}}}{{\sqrt 3 \pi {\varepsilon _0}b}}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359492
Consider a conducting spherical shell of radius \(R\). A charge \(Q\) is given onto the surface of the sphere. The total energy of the shell is
1 \(\frac{{{Q^2}}}{{4\pi {\varepsilon _0}}}\)
2 \(\frac{{{Q^2}}}{{4\pi {\varepsilon _0}R}}\)
3 \(\frac{{{Q^2}}}{{8\pi {\varepsilon _0}R}}\)
4 \(\frac{{{Q^2}}}{{\pi {\varepsilon _0}R}}\)
Explanation:
The energy of the shell is equal to the work done by an external agent to bring the charge from \(\infty \) to the surface. Let \(q\) be the charge already formed on the surface. The small work done \(d{w_{ext}}\) to bring \(dq\) from \(\infty \) to the surface is \(d{W_{ele}} = - \left( {dU} \right) = - \left( {{U_f} - {U_i}} \right)\) \( = - \left[ {\frac{{qdq}}{{4\pi {\varepsilon _0}R}} - \frac{{qdq}}{{4\pi {\varepsilon _0}\left( \infty \right)}}} \right]\) \(d{W_{ele}} = \frac{{ - qdq}}{{4\pi {\varepsilon _0}R}}\) \(d{W_{ext}} = - d{W_{ele}} = \frac{{qdq}}{{4\pi {\varepsilon _0}R}}\) \( \Rightarrow {W_{ert}} = \frac{1}{{4\pi {\varepsilon _0}R}}\int_0^Q {qdq} \) \({W_{ext}} = \frac{{{Q^2}}}{{8\pi {\varepsilon _0}R}}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359493
Three charges are placed at the vertex of an equilateral triangle as shown in figure. For what value of \(Q\) the electrostatic potential energy of the system is zero?
1 \(-q\)
2 \(q / 2\)
3 \(-2 q\)
4 \(\dfrac{-q}{2}\)
Explanation:
Electrostatic potential energy of the given system is given as. \(\begin{array}{ll}U=\dfrac{1}{4 \pi \varepsilon_{0}} \times \dfrac{1}{a}\left(Q q+Q q+q^{2}\right) \\2 Q q+q^{2}=0 & 2 Q q=-q^{2} \\2 Q=-q & Q=\dfrac{-q}{2}\end{array}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359494
Three charges \({Q,+q}\) and \({+q}\) are placed at the vertices of a right-angled isosceles triangle as shown. The net electrostatic energy of the configuration is zero if \({Q}\) is equal to
359491
If identical charges \(\left( { - q} \right)\) are placed at each corner of a cube of side \(b,\) then electric potential energy of charge \(\left( { + q} \right)\) which is placed at centre of the cube will be
Length of the longest diagonal of a cube having each side \(b\) is \(\sqrt 3 \,b.\) So distance of centre of cube from each vertex is \(\frac{{\sqrt 3 \,b}}{2}.\) Potential energy of the given system is \({U_{Sys}} = 8 \times \left\{ {\frac{1}{{4\pi {\varepsilon _0}}}.\frac{{\left( { - q} \right)\left( q \right)}}{{\sqrt 3 b/2}}} \right\} = \frac{{ - 4{q^2}}}{{\sqrt 3 \pi {\varepsilon _0}b}}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359492
Consider a conducting spherical shell of radius \(R\). A charge \(Q\) is given onto the surface of the sphere. The total energy of the shell is
1 \(\frac{{{Q^2}}}{{4\pi {\varepsilon _0}}}\)
2 \(\frac{{{Q^2}}}{{4\pi {\varepsilon _0}R}}\)
3 \(\frac{{{Q^2}}}{{8\pi {\varepsilon _0}R}}\)
4 \(\frac{{{Q^2}}}{{\pi {\varepsilon _0}R}}\)
Explanation:
The energy of the shell is equal to the work done by an external agent to bring the charge from \(\infty \) to the surface. Let \(q\) be the charge already formed on the surface. The small work done \(d{w_{ext}}\) to bring \(dq\) from \(\infty \) to the surface is \(d{W_{ele}} = - \left( {dU} \right) = - \left( {{U_f} - {U_i}} \right)\) \( = - \left[ {\frac{{qdq}}{{4\pi {\varepsilon _0}R}} - \frac{{qdq}}{{4\pi {\varepsilon _0}\left( \infty \right)}}} \right]\) \(d{W_{ele}} = \frac{{ - qdq}}{{4\pi {\varepsilon _0}R}}\) \(d{W_{ext}} = - d{W_{ele}} = \frac{{qdq}}{{4\pi {\varepsilon _0}R}}\) \( \Rightarrow {W_{ert}} = \frac{1}{{4\pi {\varepsilon _0}R}}\int_0^Q {qdq} \) \({W_{ext}} = \frac{{{Q^2}}}{{8\pi {\varepsilon _0}R}}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359493
Three charges are placed at the vertex of an equilateral triangle as shown in figure. For what value of \(Q\) the electrostatic potential energy of the system is zero?
1 \(-q\)
2 \(q / 2\)
3 \(-2 q\)
4 \(\dfrac{-q}{2}\)
Explanation:
Electrostatic potential energy of the given system is given as. \(\begin{array}{ll}U=\dfrac{1}{4 \pi \varepsilon_{0}} \times \dfrac{1}{a}\left(Q q+Q q+q^{2}\right) \\2 Q q+q^{2}=0 & 2 Q q=-q^{2} \\2 Q=-q & Q=\dfrac{-q}{2}\end{array}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359494
Three charges \({Q,+q}\) and \({+q}\) are placed at the vertices of a right-angled isosceles triangle as shown. The net electrostatic energy of the configuration is zero if \({Q}\) is equal to
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359491
If identical charges \(\left( { - q} \right)\) are placed at each corner of a cube of side \(b,\) then electric potential energy of charge \(\left( { + q} \right)\) which is placed at centre of the cube will be
Length of the longest diagonal of a cube having each side \(b\) is \(\sqrt 3 \,b.\) So distance of centre of cube from each vertex is \(\frac{{\sqrt 3 \,b}}{2}.\) Potential energy of the given system is \({U_{Sys}} = 8 \times \left\{ {\frac{1}{{4\pi {\varepsilon _0}}}.\frac{{\left( { - q} \right)\left( q \right)}}{{\sqrt 3 b/2}}} \right\} = \frac{{ - 4{q^2}}}{{\sqrt 3 \pi {\varepsilon _0}b}}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359492
Consider a conducting spherical shell of radius \(R\). A charge \(Q\) is given onto the surface of the sphere. The total energy of the shell is
1 \(\frac{{{Q^2}}}{{4\pi {\varepsilon _0}}}\)
2 \(\frac{{{Q^2}}}{{4\pi {\varepsilon _0}R}}\)
3 \(\frac{{{Q^2}}}{{8\pi {\varepsilon _0}R}}\)
4 \(\frac{{{Q^2}}}{{\pi {\varepsilon _0}R}}\)
Explanation:
The energy of the shell is equal to the work done by an external agent to bring the charge from \(\infty \) to the surface. Let \(q\) be the charge already formed on the surface. The small work done \(d{w_{ext}}\) to bring \(dq\) from \(\infty \) to the surface is \(d{W_{ele}} = - \left( {dU} \right) = - \left( {{U_f} - {U_i}} \right)\) \( = - \left[ {\frac{{qdq}}{{4\pi {\varepsilon _0}R}} - \frac{{qdq}}{{4\pi {\varepsilon _0}\left( \infty \right)}}} \right]\) \(d{W_{ele}} = \frac{{ - qdq}}{{4\pi {\varepsilon _0}R}}\) \(d{W_{ext}} = - d{W_{ele}} = \frac{{qdq}}{{4\pi {\varepsilon _0}R}}\) \( \Rightarrow {W_{ert}} = \frac{1}{{4\pi {\varepsilon _0}R}}\int_0^Q {qdq} \) \({W_{ext}} = \frac{{{Q^2}}}{{8\pi {\varepsilon _0}R}}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359493
Three charges are placed at the vertex of an equilateral triangle as shown in figure. For what value of \(Q\) the electrostatic potential energy of the system is zero?
1 \(-q\)
2 \(q / 2\)
3 \(-2 q\)
4 \(\dfrac{-q}{2}\)
Explanation:
Electrostatic potential energy of the given system is given as. \(\begin{array}{ll}U=\dfrac{1}{4 \pi \varepsilon_{0}} \times \dfrac{1}{a}\left(Q q+Q q+q^{2}\right) \\2 Q q+q^{2}=0 & 2 Q q=-q^{2} \\2 Q=-q & Q=\dfrac{-q}{2}\end{array}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359494
Three charges \({Q,+q}\) and \({+q}\) are placed at the vertices of a right-angled isosceles triangle as shown. The net electrostatic energy of the configuration is zero if \({Q}\) is equal to
