Physics NCERT Exemplar Solutions Class 12th Chapter Two Electrostatic Potential and Capacitance

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C₁ = $\frac{k1\in ?A}{d1}$  , C₂ = $\frac{k2\in ?A}{d2}$

$\frac{1}{C}=\frac{1}{C1}+\frac{1}{C2}$ = $\frac{C1\times C2}{C1+C2}$ = $\frac{k1k2\in ?A\mathrm{}}{k1d2+k2d1}$

$K$ = $\frac{k1k2(d1+d2)}{k1d2+k2d1}$

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(b), (c) We know, the electric field intensity E and electric potential V are dV related as E =- dV/dr or we can write |E|=ΔV/Δr

The electric field intensity E and electric potential V are related as E = 0 and for V = constant, dV/dr=0 this imply that electric field intensity E = 0.

If some charge is present inside the region then electric field cannot be zero at that region, for this V = constant is not valid.

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(c), (d) Case A: When key K is kept closed and plates of capacitors are moved apart using insulating handle.
The battery maintains the potential difference across connected capacitor in every circumstance. The separation between two plates increases which in turn decreases its capacitance (C=? ₀A/d)and potential difference across connected capacitor continue to be the same as capacitor is still connected with battery. So the charge stored decreases as Q = CV.

Case B: When key K is opened and plates of capacitors are moved apart using insulating handle.The charge stored by isolated charged capacitor remains conserved. The separation between two plates is increasing which in turn decreases its capacitance with the decrease of capacitance, potential difference V increases as V=Q/C.

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The force on a charge particle in electric field F = qE

The free electrons (negative charge) experience electrostatic force in a direction opposite to the direction of electric field.

The direction of electric field is always from higher potential to lower. Hence direction of travel of electrons is from lower potential to region of higher potential.

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As the electric field and force both are conservative so work done in conservative body as the body is closed is zero.

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(c) As W=qdv=q (final potential-initial potential), but the potential at a and b is same in all cases . so work done is equal in all cases.

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When initially K₁ is closed and K₂ is open, the capacitors C₁ and C₂ connected in series with battery

Q=CV=C (C₁C₂/C₁+C₂) = 6 $\times$ 3/6+3=18 $\mu$ C

Now K₁ was opened and K₂ was closed, the battery and capacitor C, are disconnected from the circuit .

Here charge being shared by both capacitor so a common potential will develop.

V=C₁V₁+C₂V₂/C₁+C₂ = 18/3+3=3V

Q₂=3CV=3 $\times$ 3=9μC

Q₃=3CV=3 $\times$ 3=9μC

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Let any three point x,y,z  seprated by a distance 2d

Then potential due to two charges

$\frac{q1}{(\sqrt[]{x^2+{\left(y\right)}^2+{(z-d)}^2}}+\frac{q2}{(\sqrt[]{x^2+{\left(y\right)}^2+{(z+d)}^2}}$ =0

$\frac{q1}{(\sqrt[]{x^2+{\left(y\right)}^2+{(z-d)}^2}}$ = $\frac{q2}{(\sqrt[]{x^2+{\left(y\right)}^2+{(z+d)}^2}}$

X²+Y²+z²+ $\frac{\left(\frac{q1}{q2}\right)+1}{\left(\frac{q1}{q2}\right)-1}$ (2zd)+d²-0

this is equation of sphere .

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$\sigma$ =charge/Area

Q= $\sigma$ (2 $\pi rdr$ )

V=KdQ/ $\surd (r$ ²+x²)=K $\sigma$ (2 $\pi rdr$ )/ $\surd (r$ ²+x²)

After integration we got

V= $k\frac{Q}{2\pi a}$ ₂ ( $\sqrt[]{x^2+a^2}$ )-x

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Surface charge density is inversely proportional to radius of the sphere so larger sphere has less charge density and vice versa.

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$\frac{1}{4\pi ??}\frac{q}{(\sqrt[]{x^2+{(d/2)}^2+h^2}}$ - $\frac{1}{4\pi ??}\frac{q}{(\sqrt[]{x^2-{(d/2)}^2+h^2}}$

If net potential is zero then,

$\frac{1}{(\sqrt[]{{\mathit{x}}^2+{(\mathit{d}/2)}^2+{\mathit{h}}^2}\mathit{}}$ = $\frac{1}{(\sqrt[]{{\mathit{x}}^2+{(\mathit{d}/2)}^2+{\mathit{h}}^2}\mathit{}}$

$x^2+{(d/2)}^2+h^2$ = $x^2-{(d/2)}^2+h^2$

2dx= 0 , x=0

X=0 , y-z plane.

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Potential energy of system of charges be algebraic sum

U = $\frac{1}{4\pi ??}(\frac{qq}{r}-\frac{qq}{r}-\frac{qq}{r})$

9 $\times$ 10⁹/10^-15 (1.6 $\times 10$ ^-19)² (1/3)²- (2/3) (1/3)- (2/3) (1/3)

-7.68 $\times$ 10^-14J

-7.68 $\times$ 10^{-14 $\times 1.6\times 10$ -19}= 0.48Mev

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To lift the disc ther must be some electrostatic force is reqired.

F=qE……. (1), E=V/d………. (2)

But the charge required during the process is - $?$ _{0 $\frac{v}{d}?$} r², using this relation in 1 and 2

F=- $?$ _{0 $\frac{v}{d}?$} r²V/d…… (3)

Also to balance something F=mg……. (4)

From equation 3 and 4

Mg= - $?$ _{0 $\frac{v}{d}?$} r²V/d, so V = $?\frac{mgd}{?0\mathrm{?}\mathrm{r}}$ ₂ this the requird solution.

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This is a long answer type question as classified in NCERT Exemplar

To lift the disc ther must be some electrostatic force is reqired.

F=qE……. (1), E=V/d………. (2)

But the charge required during the process is - $\in$ _{0 $\frac{v}{d}\pi$} r², using this relation in 1 and 2

F=- $\in$ _{0 $\frac{v}{d}\pi$} r²V/d…… (3)

Also to balance something F=mg……. (4)

From equation 3 and 4

Mg= - $\in$ _{0 $\frac{v}{d}\pi$} r²V/d, so V = $\surd \frac{mgd}{\in 0\mathrm{\pi }\mathrm{r}}$ ₂ this the requird solution.

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According to the relation that E= - $\frac{dv}{dr}$ , potential always decrease in the direction of electric field . so if we go from any point to infinity that decreases the potential.

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(c) As we know that E= - $\frac{dv}{dr}$ and potential is always same through out the surface, so according to relation electric field is zero because potential remains same at each point.

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(a) The electric potential due to point charge q is given by V=q/4? ₀r

It means electric potential due to point charge is same for all equidistant points. The locus of these equidistant points, which are at same potential, form spherical surface.

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(c) Field lines are always perpendicular to the direction of equipotential surface. The electric field in the in z- direction suggest that equipotential surface are in x-y plane. The shape of equipotential surface depends upon nature and type of distribution of charges.

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(a), (b), (c) Properties of equipotential surfaces

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(a), (d) Initially key K₁ is closed and key K₂ is open, the capacitor C₁ is charged by battery and capacitor C₂ is still uncharged. Now K₁ is opened and K₂ is closed, the capacitors C₁ and C₂ both are connected in parallel. The charge stored by capacitor C₁, gets redistributed between C₁ and C₂ till their potentials become same, i.e., V₂ = V₁.
By law of conservation of charge, the charge stored in capacitor Cx is equal to sum of charges on capacitors C₁ and C₂ when K₁ is opened and K₂ is closed, i.e.,  Q'₁+Q'₂=Q

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(a, b) The potential of a body is due to charge of the body and due to the charge of surrounding. If there are no charges anywhere else outside, then the potential of the body will be due to its own charge. If there is a cavity inside a conducting body, then charge can be placed inside the body. Hence there must be charges on its surface or inside itself.  Also The charge resides on the outer surface of a closed charged conductor. Hence there cannot be any charge in the body of the conductor.

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As we know that,   $\sigma$ =charge/Area

Q₁= $\sigma$ (4 $\pi r$ ²)

Q₂= $\sigma$ (4 $\pi \left(2r\right)$ ²)= 4Q₁

when they come in contact with each other potential being same and charge will be conserved.

So net charge is 5Q₁= 5 ( $\sigma$ (4 $\pi r$ ²)

Also V₁=V₂

KQ₁/R=KQ₂/2R  so Q₁=Q₂/2

Q₁= $\frac{5}{3}(\sigma$ (4 $\pi r$ ²) and Q₂= $\frac{10}{3}(\sigma$ (4 $\pi r$ ²)

$\sigma$ =5 $\sigma$ /3  and $\sigma$  =5 $\sigma$ /6

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Suppose the charge is slightly shifted to left so net potential energy will be

U= $\frac{1}{4\mathit{\pi }{\mathit{\epsilon }}_0}\{\frac{-{\mathit{q}}^2}{\left(\mathit{d}-\mathit{x}\right)}+\frac{-{\mathit{q}}^2}{(\mathit{d}+\mathit{x})}\}$

U= $\frac{-{\mathit{q}}^2}{4\mathit{\pi }{\mathit{\epsilon }}_{\mathit{o}}}\frac{2\mathit{d}}{({\mathit{d}}^2-{\mathit{x}}^2)}$

$\frac{\mathit{d}\mathit{U}}{\mathit{d}\mathit{x}}$ = $\frac{-{\mathit{q}}^{2}2\mathit{d}}{4\mathit{\pi }{\mathit{\epsilon }}_{\mathit{o}}}\frac{2\mathit{x}}{{({\mathit{d}}^2-{\mathit{x}}^2)}^2}$

System will be in equilibrium if this energy is zero

But when we double differentiate it

But when x<0 this value is negative so less than 0

This shows an unstable equilibrium

It follows the parabolic graph

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It is only possible when their sizes are different . when their sizes are different there capacity would change. So there is potential difference btween them.

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No, the potential function have no free space because when there is a free space there will be a leakage of potential at that point.

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The capacitance of the parallel plate capacitor, with dielectric medium of dielectric constant K is given by, C=K? ₀A/d

The capacitance of the parallel plate capacitor decreases with the removal of dielectric medium.

If we disconnect the battery from capacitor, then the charge stored will remain the same due to conservation of charge.

The energy stored in an isolated charge capacitor  U =q²/2C  as q is constant, energy stored U ∝ 1/C .As C decreases with the removal of dielectric medium, therefore energy stored increases.

The potential difference across the plates of the capacitor is given by V =q/C

Since q is constant and C decreases which in turn increases V and therefore E increases as E = V/d.

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Let us take a point p to be at a distance z from the center of ring

V=K $\int \frac{dq}{r}=k$ $\frac{dq}{\sqrt[]{z^2+a^2}}$  = $\frac{k}{\sqrt[]{z^2+a^2}}\int dq$ = $\frac{kQ}{\sqrt[]{z^2+a^2}}$

 

U= w =qV= $\frac{kQ(-q)}{\sqrt[]{z^2+a^2}}$ = $\frac{-KQq}{\sqrt[]{z^2+a^2}}$

Charge would perform oscillations.

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(d) We know that I=  $\frac{v}{R+r}$ = $\frac{2.5}{2+0.5}$ = 1 A

The potential difference across 2 ohm = 1 (2)= 2V

As Q=CV = 4 $\times$ 2= 8 μC

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If there is a potential then E? 0 electric field comes into existence, which is given by as E=-dV/dr
It means there will be field lines pointing inwards or outwards from the surface. These lines cannot be again on the surface, as the surface is equipotential. It is possible only when the other end of the field lines are originated from the charges inside. Hence, the entire volume inside must be equipotential.

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(c) Electric potential are always less in the direction of electric field

E= - $\frac{dv}{dr}$ and W= qdv . so potential energy also decrease in the direction of electric field

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Let us take a point p to be at a distance z from the center of ring

V=K $\int \frac{dq}{r}=k$ $\frac{dq}{\sqrt[]{z^2+a^2}}$  = $\frac{k}{\sqrt[]{z^2+a^2}}\int dq$ = $\frac{kQ}{\sqrt[]{z^2+a^2}}$

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(b), (c) If potential of two point is same then work done is also zero according to relation As W=qdv=q (final potential-initial potential).

Also W=-qdv= -q ${\int }_1^{2}E.dl$

For equipotential surface V₂-V₂=0 and W=0