- Home
- Standard 12
- Physics
A parallel - plate capacitor with plate area $A$ has separation $d$ between the plates. Two dielectric slabs of dielectric constant ${K}_{1}$ and ${K}_{2}$ of same area $\frac A2$ and thickness $\frac d2$ are inserted in the space between the plates. The capacitance of the capacitor will be given by :
$\frac{\varepsilon_{0} {A}}{{d}}\left(\frac{1}{2}+\frac{{K}_{1} {K}_{2}}{{K}_{1}+{K}_{2}}\right)$
$\frac{\varepsilon_{0} {A}}{{d}}\left(\frac{1}{2}+\frac{{K}_{1} {K}_{2}}{2\left({K}_{1}+{K}_{2}\right)}\right)$
$\frac{\varepsilon_{0} {A}}{{d}}\left(\frac{1}{2}+\frac{{K}_{1}+{K}_{2}}{{K}_{1} {K}_{2}}\right)$
$\frac{\varepsilon_{0} {A}}{{d}}\left(\frac{1}{2}+\frac{2\left({K}_{1}+{K}_{2}\right)}{{K}_{1} {K}_{2}}\right)$
Solution
$C_{eq}=\frac{\frac{A}{2} \varepsilon_{0}}{d}+\frac{A \varepsilon_{0}}{d} \frac{K_{1} K_{2}}{K_{1}+K_{2}}$
$=\frac{A \varepsilon_{0}}{d}\left(\frac{1}{2}+\frac{K_{1} K_{2}}{K_{1}+K_{2}}\right)$
Similar Questions
Match the pairs
| Capacitor | Capacitance |
| $(A)$ Cylindrical capacitor | $(i)$ ${4\pi { \in _0}R}$ |
| $(B)$ Spherical capacitor | $(ii)$ $\frac{{KA{ \in _0}}}{d}$ |
| $(C)$ Parallel plate capacitor having dielectric between its plates | $(iii)$ $\frac{{2\pi{ \in _0}\ell }}{{ln\left( {{r_2}/{r_1}} \right)}}$ |
| $(D)$ Isolated spherical conductor | $(iv)$ $\frac{{4\pi { \in _0}{r_1}{r_2}}}{{{r_2} – {r_1}}}$ |
