49 questions · timed · auto-graded
Two large conducting spheres carrying charges Q1 and Q2 are brought close to each other. Is the magnitude of electrostatic force between them exactly given by $\frac{\text{Q}_1\text{Q}_2}{{4}\pi\in_{0}\text{r}^{2}}$, where r is the distance between their centres?
Alternate Answer
W = q x pot. difference between the two points.
= q x 0 = 0
Alternate Answer
The work done is zero as the two points are at the same potential.
As the capacitor with the dielectric has a capacitance
$\in\text{C}$. Hence, the charge on the capacitor is given by$\text{Q}_2=\in\text{CU}=(\alpha\text{U})\text{CU}=\alpha\text{CU}$
The initial charge on the capacitor is given by
$\text{Q}_0=\text{CU}_0$
From the conservation of charges,
$\text{Q}_0=\text{Q}_1+\text{Q}_2\Rightarrow\ \text{CU}_0=\text{CU}+\alpha\text{CU}^2$
$\Rightarrow\ \alpha\text{U}_2+\text{U}-\text{U}_0=0$
$\therefore \text{U} = \frac{-1\pm\sqrt{1+4\alpha\text{U}_0}}{2\alpha}$
On solving for U0 = 78V and
$\alpha=2\text{V}^{-1}$$\text{U}=\frac{-1\pm\sqrt{1+4\times2\times78}}{2\times2}=\frac{-1\pm\sqrt{1+624}}{4}$
$=\frac{-1\pm\sqrt{625}}{4}=\frac{-1\pm25}{4}=-\frac{26}{4}\text{V and }6\text{V}$
Hence final voltage,. U = 6V.
