Electrostatic Potential and Capacitance

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

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

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

Potential energy of system of charges be algebraic sum

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

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

-7.68 $*$ 10^-14J

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

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 - $?$ _{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.

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.

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

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

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.

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

2.35 According to Gauss's law, the electric field between a sphere and a shell is determined by the charge $q_1$ on a small sphere. Hence, the potential difference, V, between the sphere and the shell is independent of charge $q_2$ . For positive charge $q_1$ , potential difference V s always positive.

2.34 Equidistant planes parallel to the x-y plane are the equipotential surfaces.

Planes parallel to the x-y plane are the equipotential surfaces with the exception that when the planes get closer, the field increases

Concentric spheres centered at the origin are equipotential surfaces.

A periodically varying shape near the given grid is the equipotential surface. This shape gradually reaches the shape of planes parallel to the grid at a larger distance.