Physics M.C.Q (1 Marks)

Using: $\text{V}=\frac{\text{W}}{\text{q}}$
$⟹ 1$ volt is equal to $1$ joule per coulomb

The thin metal plate inserted between the plates of a parallel-plate capacitor of capacitance $C$ connects the two plates of the capacitor; hence, the distance $d$ between the plates of the capacitor reduces to zero. It can be observed that the charges on the plates begin to overlap each other via the metallic plate and hence begin to conduct continuously.
Mathematically,
$\text{C}=\frac{\in_0\text{A}}{\text{d}}$
In this case,$d = 0.$
$\therefore\text{C}=\infty$

As the charge q is situated at the center of the circle $\text{ABCDE,}$ therefore the circle is an equipotential surface. That means all the points on the circle i.e. $\text{A, B, C, D,}$ and $E$ have the same potential. Therefore, work done to bring the charge from A to any point on the circle is zero always.

Electric Potential at $A$ 'due to $q ^{\prime} p V _{ A }=\frac{ Kq }{ r }$
Electric Potetial at $B$ 'due to $q^{\prime} b V_B=\frac{K q}{r}$
\& Electric potential at c 'due to $q^{\prime} p V_C=\frac{K q}{r}$
Work done $=- D _{ u }=- qDV \left\{\right.$ Let at ' P ', $\left.V _{ p }=0\right\}$
Here $V _{ A }= V _{ B }= V _{ C }$
The work done is taking a point charge from P to $A , B \& C$ is same.
So, $W _{ A }= W _{ B }= W _{ C }$

We can obtain an equivalence capacitance of $5μ\text{F},$ by connecting minimum $4$ capacitance of each $2μ\text{F},$only in the way as shown in the figure.

Total charged is conserved in the condenser. Here the potential difference on the condenser is equal to the emf of battery. When dielectric is inserted between the plates the charge will maintain the constant potential in the capacitor. Thus, the potential difference on the condenser is $4V$

The electric field in between the plates in the presence of dielectric is
$\text{E}=\frac{\text{V}}{\text{Kd}}=\frac{100}{4\times10^{-3}}=25000\text{V/m}$

Given that point $A$ is at lower electric potential than point $B$. The electron between them on line joining will move.
We have to find where this electron moves.
Since we know that electric currents move from a higher potential or a lower potential. Also, electrons move in the direction opposite to electric current. So the electron on the line joining two points $A$ and $B$ will move from lower to higher potential i.e, it will move towards $B.$

The electric field at the surface of a charged conductor is proportional to the surface charge density. The electric field is zero inside the conductor and just outside, it is normal to the surface. The contribution to the total flux comes only from its outer cross$-$section.

A charge always tries to move from a point of higher potential to a point of lower potential. The potential at A is greater than the potential at $B$ because of electric potential decreases with distance from the charge. It can also be explained by the fact that a positive charge is always repelled by another positive charge.

Energy density $U$ is given by
$\text{U}=\frac{1}{2}\in_0\text{E}^2\ \dots(1)$
The electric field created by a point charge at a distance r is given by
$\text{E}=\frac{\text{q}}{4\pi\in_0\text{r}^2}$
On putting the above form of $E$ in eq. $1$, we get
$\text{U}=\frac{1}{2}\in_0\Big(\frac{\text{q}}{4\pi\in_0\text{r}^2}\Big)^2$
Thus, U is directly proportional to $\frac{1}{\text{r}^4}.$

An electric potential $($also called the electric field potential or the electrostatic potential$)$ is the amount of electric potential energy that a unitary point electric charge would have if located at any point of space, and is equal to the work done by an electric field in carrying a unit positive charge from infinity to that point.
This value can be calculated in either a static $($time$-$invariant$)$ or a dynamic $($varying with time$)$ electric field at a specific time in units of joules per coulomb, or volts $(V).$ The electric potential at infinity is assumed to be zero.