- The electric flux is given
**D**= (2y^{2}+ z)**I**_{x}+ 4xy**I**_{y}+ x**1**_{z}C/m^{2}. The volume charge density at point (-1,0,3) isDivergence of**D**= ρ_{v}

. - Gradient of scalar field is expressed as
Gradient is operation performed on a saclar function which result in a vector function and it is the maximum rate of increase of the scalar function at a point.
- A vector field
**A**= p**1**_{n}is given in Cartesian coordinates. In cylindrical coordinates it will be represented asHints: A (r, φ, z) = (A. 1_{r}) 1_{r}+ (A. 1_{φ}) 1_{φ}+ (A. 1_{z}1_{z})

= cosφ 1_{r}+ (- sin φ) 1_{&phi}+ 0. - A potential field is given by V = 3x
^{2}y – yz. The electric field at P(2, -1, 4) isE = - divergence v

. - Which of following is zero?
The divergence of curl of a vector filed vanishes, that is

divergence of(curl of A) = 0

And also the curl of gradient of scalar field vanishes.

curl of (divergence of A) = 0. - A sphere of radius with a uniform charge density ρ
_{v}C/m^{3}shall have electric flux density at the radius r = a equal toBy Gauss’s law,

Surface integral of (D.ds) = volume integral of ρ_{v}dv)

Or, D.4π a^{2}= 4 π a^{3}ρ_{v/3.} - An infinite sheet has a charge density of 150 μ C/m. The flux density in μ C/m
^{2}isFor infinite sheet,

| E | = σ/(z ε_{o}

| D | = σ/z = 150/2 = 75 μC/m^{2}. - A parallel plate capacitor contain there dielectric layers of 1, 2 and 3 respectively and thickness of dielectric layers are 0.2 mm, 0.3 mm and 0.4 mm respectively. If area of plate is 20 cm
^{2}. What is the value of capacitance?. - An infinite non conducting sheet of charge has σ of 10
^{-7}c/m^{3}. Equipotential surface for a potential of 10 volts is.E = σ/(2 × ε_{o}

V = E.x.

C = V/E = V × ε_{o}/ σ. - If in a 1 mF capacitor, an instantaneous displacement current of 1 A is to be established in the spaces between its plates then it is possible by

. - The work done by a force
= 4u _{n}- 3u_{y}+ 2u_{z}N in giving a inc charge a displacement of (10 u_{x}+ 2u_{y - 7uz) m is 20 nj. 40 nj. 60 nj. 100 nj. Work done = qF.dl = 1(40-6-14) = 20 nj. explanation011(); } - A dipole having a moment p – 3u
_{x}- 5u_{y}+ 10u_{z}ncm is located at p(1,2,-4) in free space. The V as Q(2,3,4) is

Where D – D_{1}= Q – P = (1,1,8)

So,

. - Which statement does not say that electrostatic field conservative?
For conversation field curl of E is equal to zero.

So, E = - divergence of V and curl of ( – divergence of V) = 0. - A circular ring carrying uniformly distributed charge q and a point charge – Q on the axis of the ring. The magnitude of dipolement of the charge system is (Assume distance between centre of ring and point charge is d and radius of ring R)
For points far away, the charge on the ring may be considered to located be at the centre of the ring. Hence the dipolement becomes QD.
- If E is the electric field intensity, then what is the value of divergence of (curl of E)?.
We know that divergence of curl of any vector field is zero.
- A sphere of zoo radius contains electrical charge of density 2/(r sinθ) c/m
^{3}. What is the total charge contained within the sphere?Total charge contained within the sphere

. - Two sphere of radius r
_{1}and r_{2}are connected by a conducting wire. Each of the spheres has been given a charge Q. NowSince both of sphere has same charge. When it is connected by a conducting wire, there the potential flow higher potential to lower potential and this process stop when they have same potential. - The electrical field strength at a distance point A due to a point charge +q, located at the origin, is 100 &mul v/m. If the point charge is now enclosed by a perfectly conducting metal sheet sphere whose centre is at origin, A, outside the sphere, becomes
The charge +q induces negative charge on the inside of the sphere and positive charge on the outside. If the sphere is enclosed by a Gaussian sphere, the field strength at will be given by Gauss’s law.

This the same as through the charge is present at the origin sphere. - Plane z = 10 m carries change 20 c/m
^{2}. The electric field intensity at the origin isThe electric field due to intensity plane having charges is

E = (ρ_{o}/z ε_{o}) u_{n}

Here ρ_{o}= 10^{ - 9}/36π F/m. and u_{n}is – u(as point below the plane). - In a dielectric material an applied field in x direction E
_{x}= 5 v/m gives a polarization of p_{x}= 3/10π u_{x}nc/m^{2}. The susceptibility of the material isP = ε_{o}(ε_{r – 1}) E.

Susceptibility = ε_{r – 1}= P/(ε_{o}. E).

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