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QJEEM21Ph3aa5: JEE Main 2021 Physics - Class 12 JEEM21Ph3aa5 Solution: A parallel plate capacitor has plate area 100 m 2 and plate…

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 ε\varepsilon 0 = 8.85 ×\times 10 −- 12 F.m −- 1.

The value of 'x' to the nearest integer is _____________.

Step-by-Step Verified Solution

Area = 100 m2

Separation (d) = 10 m

Thickness = 5 m

Dielectric constant (K) = 10
JEE Main 2021 (Online) 18th March Morning Shift Physics - Capacitor Question 84 English Explanation
c1=KAε0d,c2=Aε0d{c_1} = {{KA{\varepsilon _0}} \over d},{c_2} = {{A{\varepsilon _0}} \over d}

⇒\Rightarrow ceq=c1c2c1+c2=KAε0d×Aε0dKAε0d+Aε0d{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 ceq=KA2ε02d2×dAε0(1+K){c_{eq}} = {{K{A^2}{\varepsilon _0}^2} \over {{d^2}}} \times {d \over {A{\varepsilon _0}(1 + K)}}

⇒\Rightarrow ceq=KAε0d(1+K)=10×100×8.85×10−125(1+10){c_{eq}} = {{KA{\varepsilon _0}} \over {d(1 + K)}} = {{10 \times 100 \times 8.85 \times {{10}^{ - 12}}} \over {5(1 + 10)}}

⇒\Rightarrow ceq=8.85×10−955=0.1609090×10−9{c_{eq}} = {{8.85 \times {{10}^{ - 9}}} \over {55}} = 0.1609090 \times {10^{ - 9}}

⇒\Rightarrow Ceq=160.90×10−12{C_{eq}} = 160.90 \times {10^{ - 12}}

⇒\Rightarrow Ceq=161{C_{eq}} = 161 PF

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