A) increase
B) decrease
C) remain same
D) become infinite
Correct Answer: C
Solution :
Key Idea: On inserting a thin aluminium sheet between the plates of a capacitor it act as two capacitors connected in series. The combination shows two capacitor connected in series. Resultant capacitance is \[\frac{1}{C'}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}\] \[C=\frac{K{{\varepsilon }_{0}}A}{d}\] where, A is area, d is the distance between the plates and K is dielectric constant (=1). Therefore, \[{{C}_{1}}=\frac{{{\varepsilon }_{0}}A}{d/2},{{C}_{2}}=\frac{{{\varepsilon }_{0}}A}{d/2}\] \[\therefore \] \[\frac{1}{C'}=\frac{d/2}{{{\varepsilon }_{0}}A}+\frac{d/2}{{{\varepsilon }_{0}}A}\] \[\frac{1}{C'}=\frac{d}{{{\varepsilon }_{0}}A}\] \[\Rightarrow \] \[C'=\frac{{{\varepsilon }_{0}}A}{d}=C\] Hence, on inserting aluminium sheet the capacitance remains the same.You need to login to perform this action.
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