MCQ
A parallel plate air capacitor has a capacitance $C$. When it is half filled with a dielectric of dielectric constant $5$, the percentage increase in the capacitance will be.....$\%$
- A$400$
- ✓$66.6$
- C$33.3$
- D$200$
✓
Answer
Correct option: B.
$66.6$
b
Initial capacitance $ = \frac{{{ \in _0}A}}{d}$
Initial capacitance $ = \frac{{{ \in _0}A}}{d}$
When it is half filled by a dielectric of dielectric constant $K$, then
$C_{1}=\frac{K \varepsilon_{0} A}{d / 2}=2 K \frac{\varepsilon_{0} A}{d}$
and $C_{2}=\frac{\varepsilon_{0} A}{d / 2}=\frac{2 \varepsilon_{0} A}{d}$
$\therefore \frac{1}{{{C^\prime }}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} = \frac{d}{{2{\varepsilon _0}A}}\left( {\frac{1}{K} + 1} \right)$
$ = \frac{d}{{2{\varepsilon _0}A}}\left( {\frac{1}{5} + 1} \right) = \frac{6}{{10}}\frac{d}{{{\varepsilon _0}A}}$
$C = \frac{{5{\varepsilon _0}A}}{{3d}}$
Hence, $\%$ increase in capacitance
${ = \left( {\frac{{\frac{5}{3}\frac{{{\varepsilon _0}A}}{d} - \frac{{{\varepsilon _0}A}}{d}}}{{\frac{{{\varepsilon _0}A}}{d}}}} \right) \times 100}$
${ = \left( {\frac{5}{3} - 1} \right) \times 100 = \frac{2}{3} \times 100 = 66.6\% }$
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