Physics Ncert Solutions Class 12th

This is a multiple choice answer as classified in NCERT Exemplar

(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

This is a multiple choice answer as classified in NCERT Exemplar

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

This is a multiple choice answer as classified in NCERT Exemplar

(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.

This is a multiple choice answer as classified in NCERT Exemplar

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

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

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

This is a multiple choice answer as classified in NCERT Exemplar

(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.

This is a multiple choice answer as classified in NCERT Exemplar

(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.

This is a multiple choice answer as classified in NCERT Exemplar

(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.

This is a multiple choice answer as classified in NCERT Exemplar

(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

This is a multiple choice answer as classified in NCERT Exemplar

(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 $*$ 2= 8 μC

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

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

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