- (b) $24\mu\text{J}$
Explanation:
As, $\text{C}_1=2\mu\text{F},\text{C}_2=4\mu\text{F}$
ln series combination, the equivalent capacitance will be,
$\text{C}=\frac{\text{C}_1\text{C}_2}{\text{C}_1+\text{C}_2}=\Big(\frac{2\times4}{2+4}\Big)\mu\text{F}$
$=\frac{4}{3}\mu\text{F}$
Potential difference applied, V = 6V
Energy stored in the system, $\text{U}=\frac{1}{2}\text{CV}^2$
$=\frac{1}{2}\times\frac{4}{3}\times10^{-6}\times(6)^2$
$\text{J}=24\mu\text{J}.$
- (b) $500\mu\text{J}$
Explanation:
The energy stored in a capacitor is
$\text{U}=\frac{1}{2}\text{CV}^2$
$=\frac{1}{2}\times(10\times10^{-6})(10)^2=500\mu\text{J}.$
- (b) Both V and E decrease.
Explanation:
When the gap between the plates is completely filled with dielectric of dielectric constant K, then potential is:
$\text{V}=\frac{\text{Qd}}{\text{A}\in_0\text{K}}$ (i)
and electric field is
$\text{E}=\frac{\text{Q}}{\text{A}\in_0\text{K}}$ (ii)
From equations (i) and (ii), both electric field and potential decrease.
- (b) 3.75 × 10-6J
Explanation:
Work done $=\text{U}_\text{f}-\text{U}_\text{i}=\frac{1}{2}\frac{\text{q}^2}{\text{C}_\text{f}}-\frac{1}{2}\frac{\text{q}^2}{\text{C}_\text{i}}$
$=\frac{\text{q}^2}{2}\Big[\frac{1}{\text{C}_\text{f}}-\frac{1}{\text{C}_\text{i}}\Big]$
$=\frac{(5\times10^{-6})^2}{2}\Big[\frac{1}{2\times10^{-6}}-\frac{1}{5\times10^{-6}}\Big]$
= 3.75 × 10-6J
- (b) 0.6J
Explanation:
Here r = 18cm = 18 × 10-2m, q = 5 × 10-6C
As, $\text{C}=4\pi\in_0\text{r}=\frac{18\times10^{-2}}{9\times10^9}=2\times10^{-11}\text{F}$
Energy of charged conductor is
$\text{U}=\frac{\text{q}^2}{\text{2C}}=\frac{(5\times10^{-6})^2\text{C}}{2\times2\times10^{-11}\text{F}}=0.625\text{J}.$