Physics M.C.Q (1 Marks)

The electric field due to dipole at distance $r$ is $\text{E}\propto\frac{1}{\text{r}^3}.$
Thus force on charge is $\text{F}=\text{qE}$
$\Rightarrow\text{F}\propto\frac{1}{(2\text{r})^3}=\frac{\text{F}}{8}.$

The electrostatic potential energy of the dipole is $-qLE.$
Given:
An electric dipole consisting of charges $+q$ and $-q.$
Separation between these two charges is $L.$
We need to find the electrostatic potential energy of the dipole.
We know electrostatic potential energy of the dipole is given by:
$P = -pE .....(1)$
Here, p is electric dipole defined by product of charge and distance between them.
$p = qL.$
Putting value of $p$ in equation $1,$
We get, $P = -qLE.$

Hint:- Check the dependence of capacitance on certain quantities.
In a parallel plate capacitor, the capacitance is $\text{C}=\frac{\text{k}\in_0\text{A}}{\text{d}};$
Where, $k$ is the dielectric constant, $\in_0$​ is the permittivity constant, A is the area of the conduction and $d$ is the distance between plates.
From here we can see $C$ is directly proportional to k, $\in_0$​,A and inversely proportional to d.

1. The density of the equipotential lines gives an idea about the magnitude of electric field. Higher the density, larger the field strength.
2. The direction of electric field is perpendicular to the equipotential surfaces or lines.
3. The equipotential surfaces produced by a point charge or a spherically charge distribution are a family of concentric spheres.
4. For a uniform electric field, the equipotential surfaces are a family of plane perpendicular to the field lines.
5. A metallic surface ofany shape is an equipotential surface.
6. Equipotential surfaces can never cross each other.
7. The work done in moving a charge along an equipotential surface is always zero.

The equivalent capacitance of the two $6μ\text{F} $ and $6μ\text{F} $ capacitors in series is $3μ\text{F} $.
Hence the potential across the two capacitors, original $3μ\text{F} $ capacitor and the equivalent $3μ\text{F} $ capacitor is divided equally.

When the spheres are connected, charges flow between them until they both acquire the same common potential $V.$
The final charges on the spheres are given by:
$Q _1= C _1 V \text { and } Q _2= C _2 V$
$\therefore \frac{ Q _1}{ Q _2}=\frac{ C _1 V}{ C _2 V}=\frac{ C _1}{ C _2}$

The capacity of condenser before inserting foil is $\text{C}=\frac{\text{A}\in0}{\text{d}}$​​ where A be the area of plate and d be the separation between plates.
After inserting foil the there will be two capacitors in series with capacitance $2C$ as distance is halved and the series combination of the two will give equivalent capacitance of $C,$ hence, capacity will remain same.

Here copper, tin, iron all are conductor so they will decrease the capacitance. The mica sheet is a dielectric or insulator so it will increase the capacitance $k$ times. Where $k$ is the dielectric constant.

In a parallel combination of capacitors, the potential difference across the capacitors remain the same, as the righthand-side plates and the left$-$hand$-$side plates of both the capacitors are connected to the same terminals of the battery. Therefore, the potential remains the same, that is, $V.$
For the parallel combination of capacitors, the capacitance is given by
$C_{\text {eq }}=C_1+C_2$
Here,
$C_1=C_2=C$
$\therefore C_{e q}=2 C$

Equivalent capacitance of each pair of capacitance in series$=\frac{1\times1}{1+1}\text{F}=0.5\text{F}$
The two series combination are connected in parallel. Hence the net capacitance becomes $0.5F + 0.5F = 1F$

As the capacitor is charged by using cell so potential as well as filed between the plates become constant.
For removing dielectric the capacitance becomes $C/ k$. Thus capacitance decreases by $a$ factor of $k.$

Equipotential surface
$\rightarrow P.D$ difference between two points on the surface is zero always since potential is same everywhere in equipotential surface.
$\rightarrow $ The $EF$ is always perpendicular to the surface because there is no potential gradient along any direction parallel to the surface $P$ so no $EF$ parallel to the surface
$\rightarrow $ Equipotential surface can have any shape not just sphere.
$\rightarrow $ No work is done in moving a charge along the surface, because potential difference is zero.

Equipotential surface is a surface formed by the locus of all the points which are at the same potential. Equipotential surfaces do not intersect with each other and are closely spaced in the region of strong electric fields and vice$-$versa.

If the empty Condensor has capacity $C$, then its capacity with dielectric is given by $C′ = kC$, where $k$ is the dielectric constant of the dielectric material. $k$ can never be less than $1.$

Place three $200V, 6\mu \text{F},$ capacitors in series to get $1$ equivalent $600V, 2\mu \text{F},$ capacitor. Now place $9$ of these equivalent $600V, 2\mu \text{F},$capacitors in parallel to obtain an equivalence of $18\mu F$ at $600$ Volts. All this requires a total of $27 6\mu \text{F},$ capacitors. Nine rows connected in parallel with $3$ capacitors connected in series in each row.

Consider a capacitor with charge density $σ.$
The potential between its two plates is given by $\frac{\sigma\text{d}}{\in_0}$
When a dielectric is inserted, the electric field inside the capacitor decrease decreasing the potential between the two plates of capacitor.
However, this is nothing to the dielectric strength of the dielectric.