Solution:

Negative charge.

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Solution:

Dielectrics do not have free electrons, while conductors have free electrons. When some charge is transferred to a conductor, it readily gets distributed over the entire surface of the conductor, but on insulators, the charge stays at the same place.

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Solution:

The charge on a metal spoon discharges through our body to the ground as both are conductors. But when a nylon or plastic comb is rubbed, due to the friction its acquires a negative (-ve) charge, which stays on it as it is an insulator.

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Solution:

When an object looses or gains electrons by friction/conduction/induction, then it is charged.

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Solution:

It means an electric charge is a scalar quantity and is added like algebraic numbers.

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Solution:

The minimum charge that is stable, is charge on an electron. Since, electrons are transferred from one object to another, therefore, electric charge is said to be quantised.

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Solution:

The electric force is conservative in nature, because the work done by it in moving a charge is path independent.

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Solution:

It is applied only for point charges.

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Solution:

It is a measure of the degree to which a medium can resist the movement of charges.

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Solution:

One coulomb is that charge when placed in vacuum at a distance of one metre from an equal and similar charge would repel it with a force of `9 xx 10^9 N`.

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Solution:

(a) It is conservative in nature.
(b) It depends on medium between the two charges.

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Solution:

` F = 1/(4pi K epsi_0) . (| q_1 q_2|)/r^2`

Coulomb force is inversely proportional to the dielectric constant of the intervening medium.

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Solution:

The physical quantity is an electric flux. It is a scalar quantity.

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Solution:

Dielectric constant of a medium is defined as the ratio of the force between two charges placed a certain distance apart in vacuum to the force between the same two charges placed the same distance apart in the medium. It has no units.

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Solution:

When the radius of ring is much smaller than the distance under consideration.

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Solution:

Net force experienced by any charge in a group of charges is the vector sum of the forces acting on it due to rest of the charges of the group.

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Solution:

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Solution:

Because they originate from positive ( +ve) charge and terminate at negative (-ve) charge.

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Solution:

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Solution:

From the knowledge of electric field intensity at any point, we can readily calculate the magnitude and the direction of force experienced by any charge `q_0` placed at that point.

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Solution:

Electric flux through an area is the product of magnitude of area and the component of electric field vector normal to it.

`phi_E = DeltaS ( E cos theta) = vecE . Delta vecS`

Its Sl unit is `NC^(-1) m^2`

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Solution:

The product of the magnitude of one of the point charges constituting an electric dipole and the separation between them is termed as electric dipole moment. It is a vector quantity.

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Solution:

An ideal dipole is the dipole whose size `(2a)` is vanishingly small, and the magnitude of electric charges constituting by it is very large, and the product, i.e. `2aq` is finite.

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Solution:

Since, the electric field at the location of charge `-q` is more than that of field at charge `+q` Therefore, the direction of net force will be in the direction opposite to the direction of `vecE`

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Solution:

When the dipole axis is parallel to the direction of electric field.

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Solution:

A Gaussian surface is an imaginary closed surface in three dimensional space through with the flux of a vector field is calculated.

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Solution:

A Gaussian surface is used to determine the electric field intensity around a point charge or charged body.

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Solution:

Because the electric Geld due to a system of discrete charges is not defined at the location of any charge.

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Solution:

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Solution:

Electric field intensity at a point is the electric force experienced by a unit positive charge placed at the point.

Its SI unit is `NC^(-1)` or `Vm^(-1)`

`E_p = 1/(4piepsi_0) ((2q)/a^2)`

`=> vecE_p = - vecp/(4pi epsi_0 a^3)`

It is in the direction opposite to the direction of dipole moment (i.e. from + ve to -ve charge).

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Solution:

Choose a short cylinder as a Gaussian surface about any point `P` on the surface as shown. The pill box is partly inside and partly outside the surface of the conductor. If `deltaS` = small area of cross-section, then just inside the surface, the electric field is zero; just outside, the field is normal to the surface with magnitude `E`. Thus, by Gauss's law

`E Delta S = ( sigma Delta S)/epsi_0 `

or `E = sigma/epsi_0`

or `vecE =sigma/epsi_0 hatn`

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Solution:

According to the Gauss's law, the electric flux through a dosed surface is `1/epsi_0`

times the charge enclosed by the surface.

`phi = q/epsi_0`

As the charge enclosed by the cylindrical surface is `q = lamda l`

`therefore phi = ( lamda l)/ epsi_0`

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Solution:

According to the Gauss's theorem, the total electric flux through a closed surface is `1/epsi_0` times the magnitude of net charge enclosed by the surface.

Consider a hollow charged sphere. Take a point inside the sphere where electric field is to be calculated. Draw a Gaussian surface of radius r having point P on its surface.

Now `phi = int_s vecE . vecd s = E int_s ds = E . 4 pi r^2`

According to the Gauss's theorem,

`E .4pi r^2 = q/epsi_0 \ \ \ \ \ \ \ \ ( because q = 0)`

`therefore E = 0`

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Solution:

(a) Electric flux through an area is the product of magnitude of area and the component of electric field vector normal to it.

`phi_E = DeltaS ( E costheta) = vecE Delta vecS`


Its Sl unit is `NC^(-1) m^2`


(b) Consider a thin infinite sheet of charge with uniform surface charge density `sigma`. To calculate electric field at a point `P` distant `r` from the sheet we imagine a symmetrical Gaussian surface in such a way that the point charge lies on it. Here we assume a cylinder of cross-sectional area `A` and length `2r` with its axis perpendicular to the sheet.

Flux through the curved surface of the cylinder,

`phi_1 = int vecE . vec(ds) = 0 \ \ \ \ \ \ \ \ \ \ \ ( because theta = 90^0)`

Total flux through plane faces of the cylinder,

`phi_2 = 2 int vecE . vec(ds) = 2 E A \ \ \ \ \ \ \ \ \ \ ( because theta = 0^0)`

Net flux through the Gaussian surface is

`phi = phi_1+phi_2 = 2 EA` ....................(i)

Net charge enclosed by the Gaussian surface is

`Q = sigma A`

`therefore phi = ( sigma A)/epsi_0` ...............(ii)

From equations (i) and (ii), we get

`2EA = ( sigma A) /epsi_0 => E = sigma/(2 epsi_0)`

(c) For positively charged sheet, the electric field is directed away from the sheet. For negatively charged sheet, the electric field is directed towards the plane sheet.

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Solution:

Gauss's Law states that the net outward flux through any closed surface is equal to `1/epsi_0` times the charge enclosed by the closed surface.

(i) When the point `P` is inside the shell.


In this case, the Gaussian surface lies inside the spherical shell and hence no charge is enclosed by it.

`∮ vecE . vec(ds) = ∮ E. ds cos0 = E∮ds = E xx 4pi r_1^2` ................(i)

and by Gauss's law

`∮ vecE . vec(ds) = 1/epsi_0 xx 0 = 0` ................(ii) (since no charge is enclosed)

`therefore` From (i) and (ii), we have

`E xx 4 pir_1^2 = 0`

or `E = 0`, i.e. there is no electric field inside a charged spherical shell.

(ii) When the point P lies outside the shell


Consider a spherical shell of radius `r_1` charged to a potential `V` by a charge `Q.` .To find the electric intensity at a point `P` at a distance `r_2` from the centre of the spherical shell imagine a spherical Gaussian surface to be drawn around the charged shell. At every point of this shell, the `vecE` vector and `vec(dS)` vector are directed outwards

in the same direction, i.e.` theta = 0`.

`therefore ∮ vecE . vec(ds) = ∮ E .ds = E ∮ds = E xx 4 pir_2^2` ............(i)

Also, by Gauss's law

`∮vecE . vec(ds) = 1/epsi_0 Q` ................(ii)

From (i) and (ii), we get

`E xx 4 pi r^2 = 1/epsi_0 Q => E = 1/(4pi epsi_0) Q/r^2 \ \ \ \ \ \ \ [ because r = r_2]`

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