• Equipotential surfaces are surfaces which have the same value of potential throughout their surface.
  
• For a gravitatiomal field of $$E=\dfrac{GM}{r^2}$$,the potential is $$v=-\dfrac{GM}{r}$$
• So,at a particular 'r',the potential remains constant at a value of $$-\dfrac{GM}{r}$$
• So equipotential surfaces are circles for the filed of $$E=\dfrac{GM}{r^2}$$
• Hence,the field is perpendicular to equipotential surface.

When the plate separation is double, the charge remains constant and the capacitance is halved.The Work done in separating the plate is the difference in energy stored.

Key Concept Here, we have to find the electric potential due to multiple of charges, so, we have to apply superposition principle to calculate net electric potential.
Length of the diagonals $$= \sqrt {2}\times$$ side length
$$= \sqrt {2} \times 2\sqrt {2} = 4\ m$$
$$\therefore$$ Length of semi-diagonals $$= \dfrac {4}{2} = 2m$$.
Now, potential due to charges at point $$O$$
$$V_{0} = \dfrac {!}{4\pi \epsilon_{0}} \left (\dfrac {1\times 10^{-6}}{2}\right ) \times 4$$
$$= 9\times 10^{9}\times 10^{-6} \times 2 = 18\times 10^{3}V$$.

The potential energy of system is
$$U=k\dfrac{q_1q_2}{r}$$
$$U=k\dfrac{q^2}{r}$$     $$[q_1=q_2=q]$$
or $$ U= 9\times 10^9 \times \dfrac{(2\times 10^{-6})^2}{1}$$
or $$U=9\times 10^9\times 4 \times 10^{-12}$$
or $$U=3.6 \times 10^{-2}\, J$$

The potential at the centre of the sphere is $$80$$V because it remains same at each point under the metallic hollow sphere.

Electric field due to a charge carrying hollow metallic sphere is given by
$$E_r = 0$$   for  $$r< R$$
where $$R$$ is the radius of sphere.
$$E_r = \dfrac{KQ}{r^2}$$    for  $$r\geq R$$
So, electric field inside the hollow metallic sphere is zero but non zero outside it.

Time period $$=4T(SHM$$ and is independent of d

Mass of proton $$=1$$ $$u$$

At closest distance r its whole KE is converted into PE.
$$\therefore \displaystyle\frac{1}{2}mv^2=\frac{1}{4\pi \varepsilon_0}\frac{Q\cdot q}{r}$$ $$\Rightarrow r=\displaystyle\frac{1}{4\pi \varepsilon_0}\frac{Q\cdot q}{mv^2}$$
In next case, $$\displaystyle r'=\frac{1}{4\pi \varepsilon_0}\frac{Qq}{m(2v)^2}$$
$$\Rightarrow r'=\displaystyle \frac{1}{4}\left(\displaystyle\frac{1}{4\pi \varepsilon_0}\cdot \frac{Qq}{mv^2}\right)\Rightarrow r'=r/4$$.

$$E=\cfrac{\sigma}{K\epsilon_o}$$

As dielectric slab does not affect charges:

The resultant capacitor when capacitors are joined in series is the reciprocal of the sum of the reciprocals of the indivisual capacitors.

Potential at Point $$A$$ will be given by,

$$dV=\vec E \cdot \vec dr$$

Work done by the electrostatic force is independent of the path followed by it, and it depends only on the initial and final positions. For example, work done in moving a unit positive charge in a closed loop of an electric field is zero. So,electrostatic force is a conservative force

Since all the capacitors are connected in series,
$$\therefore$$ equivalent capacitance is
$$\cfrac { 1 }{ { C }_{ eq } } =\cfrac { 1 }{ 2 } +\cfrac { 1 }{ 3 } +\cfrac { 1 }{ 6 } =1\quad \quad \therefore { C }_{ eq }=1\mu F$$
Charge on each capacitor is same
$$q={ C }_{ eq }V=1\times 10=10\mu C\quad $$ ($$\because V=10V$$)

Charge on capacitor plates without the dielectric is
$$=CV=\left( 5\times { 10 }^{ -6 }F \right) \times 1V=5\times { 10 }^{ -6 }C=5\mu C$$
The capacitance after the dielectric is introduced is
$$C'=\cfrac { { \varepsilon  }_{ 0 }A }{ d-\left( t-\cfrac { t }{ K }  \right)  } =\cfrac { { \varepsilon  }_{ 0 }A/d }{ 1-\left( \cfrac { t-\cfrac { t }{ K }  }{ d }  \right)  } =\cfrac { C }{ 1-\left( \cfrac { t-\cfrac { t }{ K }  }{ d }  \right)  } =\cfrac { 5\mu F }{ 1-\left( \cfrac { 4cm-\cfrac { 4cm }{ 4 }  }{ 6cm }  \right)  } =\cfrac { 5\mu F }{ 1-\left( \cfrac { 4-1 }{ 6 }  \right)  } =10\mu F\quad \quad $$
$$\therefore$$ Charge on capacitor plate now will be
$$Q'=C'V=10\mu F\times 1V=10\mu C$$
Additional charge transferred $$=Q'-Q=10\mu C-5\mu C=5\mu C$$

For parallel plate capacitor $$C= \varepsilon_0 \dfrac{A}{d}$$ ... 1

When the capacitor is connected to dc source, it gets charged. The battery is then disconnected, so no more charge can flow in. On removing dielectric, capacitance decreases.
Energy stored $$\left( u=\cfrac { { q }^{ 2 } }{ 2C }  \right) $$ will increase
Potential $$\left( V=\cfrac { q }{ C }  \right) $$ will also increase
Electric field $$\left( E=\cfrac { V }{ D }  \right) $$ will increase

Hint:- Check the dependence of capacitance on certain quantities.

Explanation:-

$$\bullet$$In a parallel plate capacitor, the capacitance is $$C = $$$$\dfrac{k{{\varepsilon }_{0}}A}{d}$$;

$$\bullet$$Where, k is the dielectric constant, $${{\varepsilon }_{0}}$$ is the permittivity constant, A is the area of the conduction and d is the distance between plates.

$$\bullet$$From here we can see C is directly proportional to k, $$\epsilon_o$$,A and inversely proportional to d.

⸫ So capacitance will increase with increase in Area. 

$$\textbf{Hence, the correct option is (C)}$$

When the two capacitors charged to same potential are connected in series, then total potential difference $$V'={ V }_{ 1 }+{ V }_{ 2 }=V+V=2V$$

The magnitude of the electric field due to a charged plate is given as:

Capacitance $$C=\varepsilon\dfrac{A}{d}$$ , $$A-area\ of\ the\ capacitor\ ; d-distance\ between\ the\ plates$$

$$C = \cfrac{k\epsilon_0A}{d}$$

Here $$V=\cfrac { Q }{ 4\pi { \varepsilon  }_{ 0 }r } =Q\times { 10 }^{ 11 }$$
$$\therefore 4\pi { \varepsilon  }_{ 0 }r={ 10 }^{ -11 }...(i)$$
Now, $$E=\cfrac { Q }{ 4\pi { \varepsilon  }_{ 0 }{ r }^{ 2 } } =\cfrac { Q\times 4\pi { \varepsilon  }_{ 0 } }{ (4\pi { \varepsilon  }_{ 0 }{ r) }^{ 2 } } =\cfrac { Q\times 4\pi { \varepsilon  }_{ 0 } }{ ({ 10 }^{ -11 })^{ 2 } } =4\pi { \varepsilon  }_{ 0 }Q\times { 10 }^{ 22 }V{ m }^{ -1 }$$ (Using (i))

Here $$q=5\times { 10 }^{ -7 }C,r=10cm=0.1m\quad $$
Potential, $$V=\cfrac { 1 }{ 4\pi { \varepsilon  }_{ 0 } } \cfrac { q }{ r } =\cfrac { 9\times { 10 }^{ 9 }\times 5\times { 10 }^{ -7 } }{ 0.1 } =4.5\times { 10 }^{ 4 }V$$

Since, our body and the surface of earth, both are conducting, therefore our body and the ground form an equipotential surface. As we step out in to the open from our house, the original equipotential surfaces of open air change, keeping our body and the ground at the same potential. That is why we do not get an electric shock.

$$\dfrac{{\varepsilon A}}{d} = \dfrac{{{\varepsilon _0}A}}{{d' - t + \dfrac{r}{{{\varepsilon _r}}}}}$$
or
$$d = d' - t + \dfrac{t}{{{\varepsilon _r}}}$$
$$r - \dfrac{r}{{{\varepsilon _r}}} = d' - d = 2.4\,mm$$
$$r\left( {1 - \dfrac{1}{{{\varepsilon _r}}}} \right) = 2.4$$
$$1 - \dfrac{1}{{{\varepsilon _r}}} = \dfrac{{2.4}}{3}\, \Rightarrow \,{\varepsilon _r} = \dfrac{{30}}{6} = 5$$

In this case the charge will arrange in a way that it does not create any field outside as according to Gauss law. The net charge on the system containing both plates is zero. And in between the plates

Flux $$= \dfrac{q_{in}}{\varepsilon_0}$$
The two plates of the capacitor have equal and opposite charges. Hence, net charge enclosed by the given surface $$= 0$$

$$\therefore$$Flux is zero in both cases.
Hence change in flux $$= 0$$.

In series combination of capacitor the potential across each capacitor is same and 

Work done is given difference in potentials. In an equipotential surface, all points will have same potential. Thus work done is zero

The correct option is (c)

Capacitance with air as dielectric $$C= \varepsilon_0 \dfrac{A}{d}$$

The potential is seen to be a constant on a sphere at all points. Hence (c) is the correct option

$$\vec { E } =-\dfrac{dV}{dx}$$

ref. image I  $$C = 45 \mu F$$

Two insulated charged spheres of radii $$ R_1,R_2$$having charges $$q_1,q_2$$respectively are brought in contact with each other.
Energy is decreased because system changes to get stability.

V = 1500 V