JEE MAIN - Physics (2021 - 18th March Morning Shift - No. 21)
A parallel plate capacitor has plate area 100 m2 and plate separation of 10 m. The space between the plates is filled up to a thickness 5 m with a material of dielectric constant of 10. The resultant capacitance of the system is 'x' pF.
The value of $$\varepsilon $$0 = 8.85 $$\times$$ 10$$-$$12 F.m$$-$$1.
The value of 'x' to the nearest integer is _____________.
The value of $$\varepsilon $$0 = 8.85 $$\times$$ 10$$-$$12 F.m$$-$$1.
The value of 'x' to the nearest integer is _____________.
Answer
161
Explanation
Area = 100 m2
Separation (d) = 10 m
Thickness = 5 m
Dielectric constant (K) = 10
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$${c_1} = {{KA{\varepsilon _0}} \over d},{c_2} = {{A{\varepsilon _0}} \over d}$$
$$ \Rightarrow $$ $${c_{eq}} = {{{c_1}{c_2}} \over {{c_1} + {c_2}}} = {{{{KA{\varepsilon _0}} \over d} \times {{A{\varepsilon _0}} \over d}} \over {{{KA{\varepsilon _0}} \over d} + {{A{\varepsilon _0}} \over d}}}$$
$$ \Rightarrow $$ $${c_{eq}} = {{K{A^2}{\varepsilon _0}^2} \over {{d^2}}} \times {d \over {A{\varepsilon _0}(1 + K)}}$$
$$ \Rightarrow $$ $${c_{eq}} = {{KA{\varepsilon _0}} \over {d(1 + K)}} = {{10 \times 100 \times 8.85 \times {{10}^{ - 12}}} \over {5(1 + 10)}}$$
$$ \Rightarrow $$ $${c_{eq}} = {{8.85 \times {{10}^{ - 9}}} \over {55}} = 0.1609090 \times {10^{ - 9}}$$
$$ \Rightarrow $$ $${C_{eq}} = 160.90 \times {10^{ - 12}}$$
$$ \Rightarrow $$ $${C_{eq}} = 161$$ PF
Separation (d) = 10 m
Thickness = 5 m
Dielectric constant (K) = 10
$${c_1} = {{KA{\varepsilon _0}} \over d},{c_2} = {{A{\varepsilon _0}} \over d}$$
$$ \Rightarrow $$ $${c_{eq}} = {{{c_1}{c_2}} \over {{c_1} + {c_2}}} = {{{{KA{\varepsilon _0}} \over d} \times {{A{\varepsilon _0}} \over d}} \over {{{KA{\varepsilon _0}} \over d} + {{A{\varepsilon _0}} \over d}}}$$
$$ \Rightarrow $$ $${c_{eq}} = {{K{A^2}{\varepsilon _0}^2} \over {{d^2}}} \times {d \over {A{\varepsilon _0}(1 + K)}}$$
$$ \Rightarrow $$ $${c_{eq}} = {{KA{\varepsilon _0}} \over {d(1 + K)}} = {{10 \times 100 \times 8.85 \times {{10}^{ - 12}}} \over {5(1 + 10)}}$$
$$ \Rightarrow $$ $${c_{eq}} = {{8.85 \times {{10}^{ - 9}}} \over {55}} = 0.1609090 \times {10^{ - 9}}$$
$$ \Rightarrow $$ $${C_{eq}} = 160.90 \times {10^{ - 12}}$$
$$ \Rightarrow $$ $${C_{eq}} = 161$$ PF
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