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A dielectric of dielectric constant $K$ is introduced such that half of its area of a capacitor of capacitance $C$ is occupied by it. The new capacity is
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The correct answer is:
$\frac{(1+K) C}{2}$
The dielectric is introduced such that half of its area is occupied by it.
In the given case, the two capacitors are in parallel.
$\therefore \quad C^{\prime}=C_1+C_2$
But $\quad C_1=\frac{A \varepsilon_0}{2 d}$ and $C_2=\frac{K A \varepsilon_0}{2 d}$
Thus, $\quad C^{\prime}=\frac{A \varepsilon_0}{2 d}+\frac{K A \varepsilon_0}{2 d} \Rightarrow C^{\prime}=\frac{C}{2}(1+K)$
In the given case, the two capacitors are in parallel.
$\therefore \quad C^{\prime}=C_1+C_2$
But $\quad C_1=\frac{A \varepsilon_0}{2 d}$ and $C_2=\frac{K A \varepsilon_0}{2 d}$
Thus, $\quad C^{\prime}=\frac{A \varepsilon_0}{2 d}+\frac{K A \varepsilon_0}{2 d} \Rightarrow C^{\prime}=\frac{C}{2}(1+K)$
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