
A point charge causes an electric flux of \[ - 1.0 \times {10^3}\,{\text{N}}{{\text{m}}^{\text{2}}}{{\text{C}}^{{\text{ - 1}}}}\] to pass through a spherical Gaussian surface of \[10.0\,{\text{cm}}\] radius centred on the charge.
(a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?
(b) What is the value of the point charge?
Answer
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Hint:For part (a), we need to find the flux if the radius of the Gaussian surface is doubled, for this recall the formula for electric flux, observe how the electric flux is dependent on the radius of the Gaussian surface. For part (b) use the formula for electric flux in terms of charge to the value of the given point charge.
Complete step by step answer:
Given, electric flux \[\phi = - 1.0 \times
{10^3}\,{\text{N}}{{\text{m}}^{\text{2}}}{{\text{C}}^{{\text{ - 1}}}}\]
Electric flux of a point charge can be written as,
\[\phi = \dfrac{q}{{{\varepsilon _0}}}\]
\[q\] is the charge and \[{\varepsilon _0} = 8.85 \times {10^{ -
12}}\,{\text{C}}{{\text{N}}^{{\text{ - 1}}}}{{\text{m}}^{{\text{ - 2}}}}\] is the permittivity of free space.
(a) It is asked if the radius of the Gaussian surface is doubled, what will be the value of flux.
Let \[R\] be the radius of the Gaussian surface and when it is doubled it is changed to \[2R\]. From the formula of electric flux that is \[\phi = \dfrac{q}{{{\varepsilon _0}}}\], we observe that electric flux is independent of the radius of the Gaussian surface. Therefore, the electric flux will remain same that is \[\phi = - 1.0 \times
{10^3}\,{\text{N}}{{\text{m}}^{\text{2}}}{{\text{C}}^{{\text{ - 1}}}}\].
(b) We asked to find the value of the point charge. Let \[q\] be the point charge.
Now, from the formula of electric flux we have,
\[\phi = \dfrac{q}{{{\varepsilon _0}}} \\
\Rightarrow q = \phi {\varepsilon _0} \]
Putting the value of \[\phi \] and \[{\varepsilon _0}\], we get
\[q = \left( { - 1.0 \times {{10}^3}} \right)\left( {8.85 \times {{10}^{ - 12}}} \right) \\
\therefore q = - 8.85 \times {10^{ - 9\,}}\,{\text{C}} \]
Therefore, the value of the point charge is \[ - 8.85 \times {10^{ - 9\,}}\,{\text{C}}\].
Note:Gaussian surface is applicable for point charge, for uniformly charged sphere or charge distribution with spherical symmetry. Total charge should be within the Gaussian surface. Electric flux is the measure of the electric field through the Gaussian surface. It describes the strength of the electric field.
Complete step by step answer:
Given, electric flux \[\phi = - 1.0 \times
{10^3}\,{\text{N}}{{\text{m}}^{\text{2}}}{{\text{C}}^{{\text{ - 1}}}}\]
Electric flux of a point charge can be written as,
\[\phi = \dfrac{q}{{{\varepsilon _0}}}\]
\[q\] is the charge and \[{\varepsilon _0} = 8.85 \times {10^{ -
12}}\,{\text{C}}{{\text{N}}^{{\text{ - 1}}}}{{\text{m}}^{{\text{ - 2}}}}\] is the permittivity of free space.
(a) It is asked if the radius of the Gaussian surface is doubled, what will be the value of flux.
Let \[R\] be the radius of the Gaussian surface and when it is doubled it is changed to \[2R\]. From the formula of electric flux that is \[\phi = \dfrac{q}{{{\varepsilon _0}}}\], we observe that electric flux is independent of the radius of the Gaussian surface. Therefore, the electric flux will remain same that is \[\phi = - 1.0 \times
{10^3}\,{\text{N}}{{\text{m}}^{\text{2}}}{{\text{C}}^{{\text{ - 1}}}}\].
(b) We asked to find the value of the point charge. Let \[q\] be the point charge.
Now, from the formula of electric flux we have,
\[\phi = \dfrac{q}{{{\varepsilon _0}}} \\
\Rightarrow q = \phi {\varepsilon _0} \]
Putting the value of \[\phi \] and \[{\varepsilon _0}\], we get
\[q = \left( { - 1.0 \times {{10}^3}} \right)\left( {8.85 \times {{10}^{ - 12}}} \right) \\
\therefore q = - 8.85 \times {10^{ - 9\,}}\,{\text{C}} \]
Therefore, the value of the point charge is \[ - 8.85 \times {10^{ - 9\,}}\,{\text{C}}\].
Note:Gaussian surface is applicable for point charge, for uniformly charged sphere or charge distribution with spherical symmetry. Total charge should be within the Gaussian surface. Electric flux is the measure of the electric field through the Gaussian surface. It describes the strength of the electric field.
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A point charge causes an electric flux of \[ - 1.0 \times {10^3}\,{\text{N}}{{\text{m}}^{\text{2}}}{{\text{C}}^{{\text{ - 1}}}}\] to pass through a spherical Gaussian surface of \[10.0\,{\text{cm}}\] radius centred on the charge.
(a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?
(b) What is the value of the point charge?
(a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?
(b) What is the value of the point charge?

Electric Charges & Fields Class 12 Physics - NCERT EXERCISE 1.19 | Physics NCERT | Vishal Kumar Sir
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