Physics M.C.Q (1 Marks)

Since the net charge enclosed by the Gaussian surface is zero, the total flux of the electric field through the closed Gaussian surface enclosing the capacitor is zero.
$\phi=\oint\text{E.ds}=\frac{\text{q}}{\in_0}=0$
Here,
$\phi=$ Electric flux
$q =$ Total charge enclosed by the Gaussian surface.

$B$ and $C$ points are situated on the circle, so they are on the equipotential surface. To bring a point charge from $A$ to $B$ and from $A$ to $C$ requires the same amount of work done as the initial and final points have the same potential energy in both the cases.

When the separation between two charges is increased, the electric potential Energy of charge may incease or decrease.
If Both charge are like charge then electric potential energy of charge decreases.
$\text{U}=\frac{\text{k}\text{q}_1\text{q}_2}{\text{r}}$
If Both charge are unlike charge then electric potential energy of charge increases.
$\text{U}=\frac{-\text{kq}_1\text{q}_2}{\text{r}}$

The given statement is correct. Volt or Joules per Coulomb are $SI$ unit of Potential $(V)$ i.e, work done per unit charge to move a charge from a reference point to a certain point.
And electric field $E$ is negative of the gradient of electric potential $(V)$ i.e,$\text{E}=\frac{\text{-dv}}{\text{dr}}$

$\text{V}=\frac{\text{KQ}}{\text{r}}$
$V \rightarrow $ Electric Potential

$\text{V}_\text{p}=\frac{\text{KQ}}{\text{x}}+\frac{\text{KQ}}{\text{r}-\text{x}}=\frac{\text{KQ}\text{r}}{\text{x}(\text{r}-\text{x})}$
$\frac{\text{dvp}}{\text{dx}}=\frac{-\text{kQr}(\text{r}-2\text{x})}{\big(\text{r}(\text{r}-\text{x})\big)^2}=0$

$\text{r}=2\text{x},\ \text{x}=\frac{\text{r}}{2}$
$\text{V}_\text{Pmin}=\frac{\text{KQ}}{\frac{\text{r}}{2}}+\frac{\text{KQ}}{\frac{\text{r}}{2}}=\frac{4\text{KQ}}{\text{r}}$ at $\Big(\text{x}=\frac{\text{r}}{2}\Big)$

The required voltage is $1000v$ and the capacitors are parallel as $250v.$
So number of capacitors required will be $4$ i.e $250 \times 4 = 1000$ in series.
Now example of four capacitor in series will be equal $2\mu f ($micro farade$)$ but the equivalent capacitance required is given as $16\mu f$ so there must be $8$ series of parallel arrange capacitors each of capacitor $2$ micro farad hence total number of capacitor $= 4 \times 8 = 32$

Point $A$ have positive charge.
More is the positive charge, more will be the potential. So, $A$ will be at higher potential.

Since we know that capacitance$=\frac{\text{charge}}{\text{voltage}}$
Therefore capacitance$=\frac{9}{5}=1.8\text{f.}$

As the battery remains connected across the capacitor, so the potential difference remains constant at $V_o$ even after the introduction of the dielectric slab. In this way, dielectric has an effect on potential difference.

The equipotential surface is always perpendicular to the electric field lines. In the diagram, the electric field lines are horizontal and parallel to each other. Therefore, the equipotential lines must be vertical and the points that have equal potential should be on the vertical line. Therefore, $R$ and $S$ have equal potential.

The minimum capacitance can be obtained by connecting all capacitors in series. It can be calculated as follows:
$\frac{1}{\text{C}}=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{1}{2}$
$\Rightarrow\text{C}=2\mu\text{F}$
The maximum capacitance can be obtained by connecting all capacitors in parallel. It can be calculated as follows:
$\text{C}=6+6+6=18\mu\text{F}$

The total work done $=$ energy transferred.
So, we might see the equation energy $=$ voltage $\times$ charge, $E = V \times Q,$ written as,
work $=$ voltage $\times$ charge, $W = V \times Q.$
In this case, the charge is $4 mC$, that is, $0.004 C$ and work done is $2J.$