Two oppositely charged but otherwise identical conductingplates of area 2.50 square centimeters are separated by adielectric…
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Question “Two oppositely charged but otherwise identical conductingplates of area 2.50 square centimeters are separated by adielectric…”
Answer
This problem can be solved using the Capacitance of an capacitor with dielectric.
Calculating the charge density is used to calculate the magnitude of the charge per unit area of the conducting plate.
Calculating the charge density of conducting plates is used to calculate the magnitude of charge per unit area of dielectric plate.
The expression for energy stored within the capacitor calculates the total electric-field energy in the capacitor.
The charge per unit area of the conducting plate is:
\sigma = k\varepsilon _0EHere, k
For the area of dielectric plates, the magnitude of charge per unit area is:
\sigma ' = \sigma \left( 1 - \frac1k \right)Here, \sigma
To calculate the energy total stored in the capacitor, use the following expression:
U = \frac{1}{2}k{\varepsilon _0}{E^2}AdHere, k
(a)
Calculate the area of the conducting plate that is charged per unit area.
The charge per unit area of the conducting plate is:
\sigma = k\varepsilon _0EHere, k
Substitute 3.60 to k, 8.85 \times {10^{ - 12}}\,{{\rm{C}}^{\rm{2}}}{\rm{/N}} \cdot {{\rm{m}}^{\rm{2}}} \beginarrayc\\\sigma = \left( 3.60 \right)\left( 8.85 \times 10^ - 12\,\rmC^\rm2\rm/N \cdot \rmm^\rm2 \right)\left( 1.20 \times 10^6\,\rmV/m \right)\\\\ = 3.82 \times 10^ - 5\,\,\rmC^\rm2\rm/N \cdot \rmm^\rm2\\\endarray
(b)
Calculate the charge per unit area of the dielectric plate.
For the area of the dielectric plate’s surface, the magnitude of charge per unit area is:
\sigma ' = \sigma \left( 1 - \frac1k \right)Here, \sigma
Substitute 3.82 \times {10^{ - 5}}\,{{\rm{C}}^{\rm{2}}}{\rm{/N}} \cdot {{\rm{m}}^{\rm{2}}} \begin{array}{c}\\\sigma ' = 3.82 \times {10^{ - 5}}\,{{\rm{C}}^{\rm{2}}}{\rm{/N}} \cdot {{\rm{m}}^{\rm{2}}}\left( {1 - \frac{1}{{3.60}}} \right)\\\\ = 2.76 \times {10^{ - 5}}\,{\rm{C/}}{{\rm{m}}^{\rm{2}}}\\\end{array}
(c)
Use the following expression to calculate the total electric-field energy contained in the capacitor:
U = \frac{1}{2}k{\varepsilon _0}{E^2}AdMultiply the {10^{ - 4}} to convert the area from a centimeter square into a meter square.
\begin{array}{c}\\A = \left( {2.5\,\,{\rm{c}}{{\rm{m}}^{\rm{2}}}} \right)\left( {\frac{{{{10}^{ - 4}}\,{{\rm{m}}^{\rm{2}}}}}{{1\,\,{\rm{c}}{{\rm{m}}^{\rm{2}}}}}} \right)\\\\ = 2.5 \times {10^{ - 4}}\,\,{{\rm{m}}^{\rm{2}}}\\\end{array}Use {10^{ - 3}} to convert the diameter from millimeters to meters.
\begin{array}{c}\\d = \left( {1.8\,{\rm{mm}}} \right)\left( {\frac{{{{10}^{ - 3}}\,{\rm{m}}}}{{1\,\,{\rm{mm}}}}} \right)\\\\ = 1.8 \times {10^{ - 3}}\,{\rm{m}}\\\end{array}Substitute 3.60 to k, 8.85 \times {10^{ - 12}}\,{{\rm{C}}^{\rm{2}}}{\rm{/N}} \cdot {{\rm{m}}^{\rm{2}}}
\begin{array}{c}\\U = \frac{1}{2}\left( {3.60} \right)\left( {8.85 \times {{10}^{ - 12}}\,{{\rm{C}}^{\rm{2}}}{\rm{/N}} \cdot {{\rm{m}}^{\rm{2}}}} \right){\left( {1.20 \times {{10}^6}\,{\rm{V/m}}} \right)^2}\left( {2.5 \times {{10}^{ - 4}}\,{{\rm{m}}^{\rm{2}}}} \right)\left( {1.8 \times {{10}^{ - 3}}\,{\rm{m}}} \right)\\\\ = 1.03 \times {10^{ - 5}}\,{\rm{J}}\\\end{array}Ans – Part a
CodeMathTag12Part b is the area of conductor plate that has the greatest charge per unit area
CodeMathTag22Part C is the charge density per area of dielectric plates.
CodeMathTag23 is the total electric field-energy that has been stored in the capacitor
Conclusion
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