
The minimum surface charge density on the plate, so that a body of mass \[2\,{\text{kg/}}{{\text{m}}^{\text{2}}}\] may just be lifted, is
A. \[2.84 \times {10^{ - 5}}\,{\text{C/}}{{\text{m}}^{\text{2}}}\]
B. \[2.25 \times {10^{ - 5}}\,{\text{C/}}{{\text{m}}^{\text{2}}}\]
C. \[1.86 \times {10^{ - 5}}\,{\text{C/}}{{\text{m}}^{\text{2}}}\]
D. None of these
Answer
486.6k+ views
Hint: Use the formula for the force per unit area of a charged conductor. Use the concept that this force per unit area of the plate is balanced by the weight per unit area of the body. This gives the relation between the surface charge density on the plate, mass per unit area of the body and permittivity of free space.
Formula used:
The force per unit area \[\dfrac{F}{{ds}}\] of a charged conductor is given by
\[\dfrac{F}{{ds}} = \dfrac{{{\sigma ^2}}}{{2{\varepsilon _0}}}\] …… (1)
Here, \[\sigma \] is the surface charge density and \[{\varepsilon _0}\] is the permittivity of free space.
Complete answer: or Complete step by step answer:
The mass per unit area of the body is \[2\,{\text{kg/}}{{\text{m}}^{\text{2}}}\] and the body is just lifted.
\[m = 2\,{\text{kg/}}{{\text{m}}^{\text{2}}}\]
Hence, the surface per unit area \[\dfrac{F}{{ds}}\] of the plate is balanced by the weight per unit area \[mg\] of the body.
\[\dfrac{F}{{ds}} = mg\]
Here, is the mass per unit area of the body.
The permittivity of free space \[{\varepsilon _0}\] is \[8.85 \times {10^{ - 12}}\,{{\text{C}}^2}/{\text{N}} \cdot {{\text{m}}^2}\].
\[{\varepsilon _0} = 8.85 \times {10^{ - 12}}\,{{\text{C}}^2}/{\text{N}} \cdot {{\text{m}}^2}\]
Determine the surface charge density on the plate.
Substitute for \[\dfrac{F}{{ds}}\] in the above equation.
\[\dfrac{{{\sigma ^2}}}{{2{\varepsilon _0}}} = mg\]
Rearrange the above equation for the square of surface charge density \[{\sigma ^2}\] on the plate.
\[{\sigma ^2} = 2{\varepsilon _0}mg\]
Take square root on both sides of the above equation.
\[\sigma = \sqrt {2{\varepsilon _0}mg} \]
Substitute \[8.85 \times {10^{ - 12}}\,{{\text{C}}^2}/{\text{N}} \cdot {{\text{m}}^2}\] for \[{\varepsilon _0}\], \[2\,{\text{kg/}}{{\text{m}}^{\text{2}}}\] for \[m\] and \[9.8\,{\text{m/}}{{\text{s}}^2}\] for \[g\] in the above equation.
\[\sigma = \sqrt {2\left( {8.85 \times {{10}^{ - 12}}\,{{\text{C}}^2}/{\text{N}} \cdot {{\text{m}}^2}} \right)\left( {2\,{\text{kg/}}{{\text{m}}^{\text{2}}}} \right)\left( {9.8\,{\text{m/}}{{\text{s}}^2}} \right)} \]
\[ \Rightarrow \sigma = \sqrt {346.92 \times {{10}^{ - 12}}} \]
\[ \Rightarrow \sigma = 18.62 \times {10^{ - 6}}\]
\[ \Rightarrow \sigma = 1.86 \times {10^{ - 5}}\,{\text{C/}}{{\text{m}}^{\text{2}}}\]
Therefore, the surface charge density on the plate is \[1.86 \times {10^{ - 5}}\,{\text{C/}}{{\text{m}}^{\text{2}}}\].
So, the correct answer is “Option C”.
Note:
Generally, force on an object is balanced by its weight. Here, the force per unit area is balanced by weight per unit area as the area on both sides of the equation gets cancelled.
Formula used:
The force per unit area \[\dfrac{F}{{ds}}\] of a charged conductor is given by
\[\dfrac{F}{{ds}} = \dfrac{{{\sigma ^2}}}{{2{\varepsilon _0}}}\] …… (1)
Here, \[\sigma \] is the surface charge density and \[{\varepsilon _0}\] is the permittivity of free space.
Complete answer: or Complete step by step answer:
The mass per unit area of the body is \[2\,{\text{kg/}}{{\text{m}}^{\text{2}}}\] and the body is just lifted.
\[m = 2\,{\text{kg/}}{{\text{m}}^{\text{2}}}\]
Hence, the surface per unit area \[\dfrac{F}{{ds}}\] of the plate is balanced by the weight per unit area \[mg\] of the body.
\[\dfrac{F}{{ds}} = mg\]
Here, is the mass per unit area of the body.
The permittivity of free space \[{\varepsilon _0}\] is \[8.85 \times {10^{ - 12}}\,{{\text{C}}^2}/{\text{N}} \cdot {{\text{m}}^2}\].
\[{\varepsilon _0} = 8.85 \times {10^{ - 12}}\,{{\text{C}}^2}/{\text{N}} \cdot {{\text{m}}^2}\]
Determine the surface charge density on the plate.
Substitute for \[\dfrac{F}{{ds}}\] in the above equation.
\[\dfrac{{{\sigma ^2}}}{{2{\varepsilon _0}}} = mg\]
Rearrange the above equation for the square of surface charge density \[{\sigma ^2}\] on the plate.
\[{\sigma ^2} = 2{\varepsilon _0}mg\]
Take square root on both sides of the above equation.
\[\sigma = \sqrt {2{\varepsilon _0}mg} \]
Substitute \[8.85 \times {10^{ - 12}}\,{{\text{C}}^2}/{\text{N}} \cdot {{\text{m}}^2}\] for \[{\varepsilon _0}\], \[2\,{\text{kg/}}{{\text{m}}^{\text{2}}}\] for \[m\] and \[9.8\,{\text{m/}}{{\text{s}}^2}\] for \[g\] in the above equation.
\[\sigma = \sqrt {2\left( {8.85 \times {{10}^{ - 12}}\,{{\text{C}}^2}/{\text{N}} \cdot {{\text{m}}^2}} \right)\left( {2\,{\text{kg/}}{{\text{m}}^{\text{2}}}} \right)\left( {9.8\,{\text{m/}}{{\text{s}}^2}} \right)} \]
\[ \Rightarrow \sigma = \sqrt {346.92 \times {{10}^{ - 12}}} \]
\[ \Rightarrow \sigma = 18.62 \times {10^{ - 6}}\]
\[ \Rightarrow \sigma = 1.86 \times {10^{ - 5}}\,{\text{C/}}{{\text{m}}^{\text{2}}}\]
Therefore, the surface charge density on the plate is \[1.86 \times {10^{ - 5}}\,{\text{C/}}{{\text{m}}^{\text{2}}}\].
So, the correct answer is “Option C”.
Note:
Generally, force on an object is balanced by its weight. Here, the force per unit area is balanced by weight per unit area as the area on both sides of the equation gets cancelled.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Express the following as a fraction and simplify a class 7 maths CBSE

Trending doubts
Give 10 examples of unisexual and bisexual flowers

Draw a labelled sketch of the human eye class 12 physics CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE

What are the major means of transport Explain each class 12 social science CBSE

Franz thinks Will they make them sing in German even class 12 english CBSE
