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New answer posted
a year agoContributor-Level 10
This is a multiple choice answer as classified in NCERT Exemplar
(a), (b), (c) Properties of equipotential surfaces
New answer posted
a year agoContributor-Level 10
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.
New answer posted
a year agoContributor-Level 10
This is a multiple choice answer as classified in NCERT Exemplar
C1 = , C2 =
= =
=
New answer posted
a year agoContributor-Level 10
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.
New answer posted
a year agoContributor-Level 10
This is a multiple choice answer as classified in NCERT Exemplar
(c) As we know that E= - and potential is always same through out the surface, so according to relation electric field is zero because potential remains same at each point.
New answer posted
a year agoContributor-Level 10
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.
New answer posted
a year agoContributor-Level 10
This is a multiple choice answer as classified in NCERT Exemplar
(c) Electric potential are always less in the direction of electric field
E= - and W= qdv . so potential energy also decrease in the direction of electric field
New answer posted
a year agoContributor-Level 10
This is a multiple choice answer as classified in NCERT Exemplar
(d) We know that I= = = 1 A
The potential difference across 2 ohm = 1 (2)= 2V
As Q=CV = 4 2= 8 μC
New answer posted
a year agoContributor-Level 10
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 = =

New answer posted
a year agoContributor-Level 10
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 = =
U= w =qV= =
Charge would perform oscillations.

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