A) \[Q=\frac{{{\varepsilon }_{0}}AV}{d}\]
B) \[W=\frac{{{\varepsilon }_{0}}A{{V}^{2}}}{2kd}\]
C) \[E=\frac{V}{kd}\]
D) \[W=\frac{{{\varepsilon }_{0}}A{{V}^{2}}}{2d}\left( 1-\frac{1}{k} \right)\]
Correct Answer: B
Solution :
After inserting the dielectric slab New capacitance \[C'=K.C=\frac{K{{\varepsilon }_{0}}A}{d}\] New potential difference \[V'=\frac{V}{K}\] New charge \[Q'=C'V'=\frac{{{\varepsilon }_{0}}AV}{d}\] New electric field \[E'=\frac{V'}{d}=\frac{V}{Kd}\] Work done (W) = Final energy ? Initial energy W \[=\frac{1}{2}C'V{{'}^{\,2}}-\frac{1}{2}C{{V}^{2}}\] \[=\frac{1}{2}(KC)\,{{\left( \frac{V}{K} \right)}^{2}}-\frac{1}{2}C{{V}^{2}}\] \[=\frac{1}{2}C{{V}^{2}}\left( \frac{1}{K}-1 \right)=-\frac{1}{2}C{{V}^{2}}\left( 1-\frac{1}{K} \right)\] \[=-\frac{{{\varepsilon }_{0}}A{{V}^{2}}}{2d}\left( 1-\frac{1}{K} \right)\] so \[|W|\,=\frac{{{\varepsilon }_{0}}A{{V}^{2}}}{2d}\left( 1-\frac{1}{K} \right)\].You need to login to perform this action.
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