Electrostatic Potential and Capacitance

2.16 Electric field on one side of a charged body is $E_1$ and electric field on the other side of the same body is $E_2.$ If infinite plane charged body has a uniform thickness, then electric field due to one surface of the charged body is given by,

$\stackrel{?}{E_1}$ = $\frac{\sigma }{2{\epsilon }_0}\stackrel{?}{n}$ ………….(i)

Where,

$\stackrel{?}{n}$ = unit vector normal to the surface at a point

$\sigma$ = surface charge density at that point

Electric field due to the other surface of the charged body is given by

$\stackrel{?}{E_2}$ = $\frac{\sigma }{2{\epsilon }_0}\stackrel{?}{n}$ ………….(ii)

$\mathrm{E}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{c}\mathrm{}\mathrm{f}\mathrm{i}\mathrm{e}\mathrm{l}\mathrm{d}\mathrm{}\mathrm{a}\mathrm{t}\mathrm{}\mathrm{a}\mathrm{n}\mathrm{y}\mathrm{}\mathrm{p}\mathrm{o}\mathrm{i}\mathrm{n}\mathrm{t}$ due to the two surfaces,

$\stackrel{?}{E_2}-\stackrel{?}{E_1}=\frac{\sigma }{2{\epsilon }_0}\stackrel{?}{n}+\frac{\sigma }{2{\epsilon }_0}\stackrel{?}{n}=\frac{\sigma }{{\epsilon }_0}\stackrel{?}{n}$ ……(iii)

$\mathrm{S}\mathrm{i}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{s}\mathrm{i}\mathrm{d}\mathrm{e}\mathrm{}\mathrm{a}\mathrm{}\mathrm{c}\mathrm{l}\mathrm{o}\mathrm{s}\mathrm{e}\mathrm{d}\mathrm{}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r},\mathrm{}\stackrel{?}{{\mathrm{E}}_1}=0,$

$\stackrel{?}{E}$ = $\stackrel{?}{E_2}$ = $\frac{\sigma }{{\epsilon }_0}\stackrel{?}{n}$

Therefore, the electric field just outside the conductor is $\frac{\sigma }{{\epsilon }_0}\stackrel{?}{n}$

When a charged particle is moved from one point to the other on a closed loop, the work done by the electrostatic field is zero. Hence, the tangential component of electrostatic field is continuous from one side of a charged surface to the other.

2.31 The force between two conducting spheres is not exactly given by the expression ${\mathit{Q}}_1{\mathit{Q}}_2$ /4 $\mathit{?}{\mathit{?}}_0{\mathit{r}}^2$ , because there is non-uniform charge distribution on the spheres.

Gauss's law will not be true, if Coulomb's law involved $\frac{1}{r^3}$ dependence, instead of $\frac{1}{r^2}$ , on r

Yes. If a small test charge is released at rest at a point in an electrostatic field configuration, then it will travel along the field lines passing through the point, only if the field lines are straight. This is because the field lines give the direction of acceleration and not the velocity.

Whenever the electron completes an orbit, either circular or elliptical, the work done by the field of a nucleus is zero.

No. Electric field is discontinuous across the surface of a charged conductor. However, electric potential is continuous.

The capacitance of a single conductor is considered as a parallel plate capacitor with one of its two plates at infinity.

Water has an unsymmetrical space as compared to mica. Since it has a permanent dipole moment, it has a greater dielectric constant than mica.

This is a multiple choice answer as classified in NCERT Exemplar

(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.

This is a multiple choice answer as classified in NCERT Exemplar

(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