#### The number of electrons in 1 μC charge is ____ .

Charge q is given by q = ne, where q is the number of electrons and e is the basic unit of charge, which is equal to 1.6 × 10^{-19} C

Therefore, by substituting the values,

1 × 10^{-6} = n × 1.6 × 10^{-19}

n = 10^{-6}/1.6×10^{-19}

n = 10^{13}/1.6⇒n = 10^{14}/16 = 625 × 10^{10}

n = 6.25 × 10^{12}

#### A positively charged glass rod ____ .

A positively charged rod can attract not only negatively charged or less positively charged objects; it can also attract neutral objects. For example, if a positively charged rod is brought near a neutral paper, the rod exerts attraction, because, by induction, the rod makes the near end of the paper to have more electrons and hence attracting it.

#### If the electric field at a distance of 5 cm from an infinite metal sheet with a surface charge density, σ is E. Then the electric field at a distance of 10 cm from the same metal sheet is ____.

The electric field due to a long charged metal sheet is, E=σ/2ϵ0 But the electric field near the plate does not depend on the distance. Thus the electric field at a distance of 5 cm is E and at a distance of 10 cm is also E.

#### An electrified glass rod is brought near a non-conducting light object. The rod induces ____charges on the near surface and ____ charges move to the farther side. The mass of the rod will ____.

When an electrified glass rod is brought near a light object, the rod induces opposite charges on the near surface and hence similar charges move farther. This happens even if the light object is not a conductor. Contrary to the process of charging by contact, the charged glass rod does not lose or gain any of its charge, and thus the mass will remain unchanged.

#### Ironed thin strips of paper get attracted to a TV screen because ____.

Ironed thin strips of paper get attracted to a TV screen because charged paper strips attract the opposite charge on the TV screen. When a strip of paper is "hot" ironed, it acquires charge. When this is brought close to the TV screen, due to difference in the polarities of the paper strip and that on the screen, paper strip gets attracted to the screen.

#### The falling of a water droplet of mass 1 mg is just prevented by upward electric field of magnitude 0.1 kN/C. The charge on the droplet of water is ____. [Consider g = 9.8 m/s^{2}]

Mass of the water droplet = m = 1 mg = 10^{-6} kg

Electric field, E = 0.1 kN/C

= 0.1 × 10^{6} N/C

Therefore, E = 10^{5} N/C

The forces that act on the droplet are:

(a) The force due to gravity, which is mg = F_{1}

(b) The electric force, F_{2}

We know that, F_{2} = qE

For the droplet to be just prevented from falling, the two forces F_{1 }and F_{2} should be equal.

∴ F_{1 }= F_{2}

⇒ mg = qE

⇒10^{- 6}×9. 8 = q×10^{5}

⇒q = 9. 8×10^{- 6}/10^{5 }= 9. 8×10^{- 11}C

∴ q = 9.8 × 10^{-11} C

#### When a body is ____charged, its mass ____.

When a body is negatively charged, its mass increases.
An electron's mass is about 9.11 × 10^{-31} kg.
When a body is positively charged, it has lost electrons and so the mass decreases.
On the other hand, if a body is negatively charged, it gains electrons and so the mass increases.

#### A charge 'q' is to be divided between two objects A and B. The charges on A and B such that the force between A and B can be maximum is ____.

Let the two objects A and B hold the charges 'q_{1}' and q - q_{1} respectively.

Then, the force between A and B is:

F = 1/4Πε_{0}[(q_{1})(q -q_{1})/r^{2}], where r is the distance between A and B.

F = k[q_{1}(q -q_{1})/r^{2 }] ------------------ (1) k = 1/4Πε0r^{2 }a constant

For 'F' to be maximum, dF/dq_{1} = 0.

Therefore, differentiating equation 1 w.r.t q_{1},

dF/dq_{1 }= k d/dq_{1 }[q_{1}(q -q_{1})] = k (q - 2q_{1})

∴dF/dq_{1} = 0⇒k (q - 2q_{1})= 0⇒ q = 2q_{1}⇒q_{1}= q/2

#### The electric field strength at a point on the axis of an electric dipole is ____ the field strength at the same distance on the equatorial line. (Assume r > > a, where r is the axial/equatorial point distance, 2a is the length of dipole)

Dipole electric field at an axial point at a distance 'r' from the centre of the dipole is

E_{a}=1/4πε_{0}×2pr^{3}(>>; a)

The electric field due to a dipole at an equatorial point at distance r from the centre of the dipole is

E_{e}=1/4πε_{0}×p/r^{3} (r >>; a)

∴E_{a}/E_{e}=2/1⇒E_{a}= 2E_{e}

#### Gauss' law is valid for ____.

Gauss' law is applicable for any closed surface irrespective of regular or irregular surfaces and sizes. But it is not applicable for open surfaces.

#### For a point outside the shell, the electric field due to uniformly charged spherical shell is similar to the electric field due to ___.

For a point outside the shell, the electric field due to uniformly charged spherical shell is similar to the electric field due to a point charge placed at the centre of the shell, which is equal to q/4πε_{0}r^{2}

#### An electric dipole formed by charge +6 μC and -6 μC, separated by a distance of 2 mm. The dipole moment is ____ C m, directed from____ to ____.

The dipole moment is given by p = q × 2a, where q is the charge and 2a is the separation distance.

∴ p = 6 × 10^{-6} × 2 × 10^{-3} Cm (q = 6 μC; 2a = 2 mm)

⇒ p = 12 × 10^{-9} C m

#### A thin metal sheet carrying a positive charge of surface charge density 1.77××10^{−8}Cm^{−2}.The electric field at a distance of 8 cm from the plate is ____.

Given,
Surface charge density, σ=1.77××10^{−8}Cm^{−2}

The distance from the plate, r=8cm

But the electric field due to a charged metal sheet is, E=σ/2ϵ_{0}

=1.77×10^{−8 }/2×8.85×10^{−12}

=10^{3}NC^{−1}

#### If a cubic conductor of side 20 cm is charged to 15 µC, the surface charge density acquired is ____ C/m^{2}

Here, surface area = 6a^{2} = 6 × (0.2)^{2} m^{2}

Charge = q = 15µC

Therefore, Surface charge density = q/6a^{2}C/m^{2}

=15×10^{−6 }/6×0.2×0.2C/m^{2}

= 0.625 × 10^{-4} C/m^{2}

= 6.25 × 10^{-5} C/m^{2}

#### As the distance from the infinitely long charged wire increases, the strength of the electric field ____.

We know that.

The electric field due to an infinitely long charged wire is, E=λ/2πϵ_{0}r

Where λ is the linear charge density, which is constant for the wire.

Then, E∝1/r

As the distance from the wire increase, the strength of the electric field decreases.

#### When an Ebonite rod is rubbed with fur, the charge acquired by fur is ____and that by Ebonite is ____.

The electrons of the outer shell of an atom are not tightly bound to the nucleus. When two different bodies are rubbed against each other, these electrons are transferred from one body to the other. In this, case, when an Ebonite rod is rubbed with fur, electrons from the fur are transferred to the rod. Thus, by giving electrons, fur acquires a positive charge, whereas, by taking electrons, the rod acquires a negative charge.

#### Gauss' law is true only if force due to a charge varies as ____.

Gauss' law is based on the inverse square dependence on distance. Any violation of Gauss' law will indicate departure from inverse square law.

#### The unit of dipole moment is ____.

Dipole moment is given by P = q × 2a

q has the unit 'C'.

2a, which is the distance of the separation of the dipole has the unit 'm'

Therefore, the unit of dipole moment = Cm

#### When a metal rod is held in hand and rubbed with wool, it will ____.

A metal rod is held in hand. As human body is a good conductor, the charge developed due to rubbing, leak through the hand to the ground. Hence, the metal rod will not be charged.

#### Which of the following statement is true ____.

Option "D" is the right option, as "electrifying" or charging involves transfer of electrons (either losing or gaining).

Option "A" is wrong as charge is not created by rubbing; only transfer of electrons happen.

Option "B" is wrong; as losing electrons is equivalent to saying the body becomes positively charged.

Option "C" is wrong, as electrostatics deals with objects in rest only.

#### An infinitely long wire carries a positive charge of linear density λ. The direction of force acting on a unit positive charge placed at a distance r from the wire is ____.

The electric field direction at a point in a plane normal to the wire is inwards, if the linear charge density (λ) is negative. If the linear charge density (λ) is positive, the direction of electric field is outwards. The direction of force acting on a positive charge placed in an external electric field is in the direction of electric field. Thus the force acting on the unit positive charge placed near the infinitely charged wire is outwards.

#### There are four charged bodies A, B, C, D.

Given, body A is negatively charged.

B attracts A. Hence, "B" is positively charged.

As "B" repels "C", "C" is also positively charged.

Also, as "D" repels "A", it should be of the same charge as that of A, which is negative.

#### If the linear charge density, λ of an infinitely long straight wire is negative, the direction of electric field, E at a certain distance in a plane normal to the wire is ____.

The electric field direction at a point in a plane normal to the wire is inwards, if the linear charge density-λ is negative. If the linear charge density-λ is positive, the direction of electric field is outwards.

#### The inward electric flux due to an electron placed in air is ____.

Let us say the charge of an electron as e

According to Gauss' law, Φ=q/ϵ_{0}

Then the inward electric flux due to the electron is, Φ=e/ϵ_{0}

#### The unit for torque experienced by an electric dipole is ____.

Torque experience by an electric dipole in a uniform external field is given by τ= p E sinθ.

Unit of p, the dipole moment is Cm.

Unit of E, the electric field strength is N/C.

Therefore, unit of τ is Cm × N/C i.e. Nm.

#### To get a surface charge density of 1/2π Cm^{−2},a sphere of radius 10cm should possess a charge of ____ m C.

In case of a spherical body, the surface charge density = σ = q/4πr^{2 }

⇒q=σ×4πr^{2}

=1/2π×4π×(0.1)^{2} (because r = 10 cm = 0.1 m)

= 2 × (0.1)^{2}

⇒ q = 0.02 C

= 20 mC

Thus, the sphere should possess 20 mC of charge.

#### Electric field lines ____.

Electric field lines extend infinitely long, but the electric field strength will be low at longer distances from the charge. Electric field lines point radially outward in positive charge, but inward in negative charge. Thus, 'B' is wrong. Option 'D' is wrong, as closely packed lines indicate stronger electric field lines. Option 'C' is right. Electric field lines are imaginary and the electric field intensity is the actual phenomenon that is explained by the lines.

#### Which of the following statements is false ____.

Field lines do not intersect each other. If they do so, there will be two directions for the electric field, which is impossible. Thus, option A is the statement that is false. Options B, C, D are properties of electric field lines.

#### The magnitude of electric field 'E' due to a point charge has ____.

The magnitude of the electric field E due to charge Q depends only on the distance 'r' of the test charge 'q' from 'Q'. Therefore, at equal distances from charge Q, the magnitude of its electric field is the same. The magnitude of E due to a point charge is thus, the same on a sphere with the point charge at the centre. Thus, it has a spherical symmetry.

#### A thin spherical shell of radius, R carries a positive charge of surface charge density, σ. Then the electric field at a distance, x (x>R) from the centre of the spherical shell is proportional to ____.

We know that,

The electric field due to a charged spherical shell at a point outside the spherical shell is, E=q/4πε_{0}r^{2}

The charge on the shell and the permittivity of air are constants.

Then E∝1/r^{2}

In this case, the distance is given as x, Then E=1/x^{2}

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