Physics ELECTRIC CHARGES AND FIELDS

### Definition Based Problems -Set 1

Set 1 - 1 , 2, 3 marks Problem
Q 3211291120

One end of a copper wire is connected to a neutral pith ball and other end to a negatively charged plastic rod What will be the charge acquired by a pith ball?

Solution:

Negative charge.
Q 3281467327

Comparison between an insulator (dielectric) and a conductor.

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.
Q 3221291121

Why does a nylon or plastic comb get electrified on combing or rubbing but a metal spoon does not?

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.
Q 3201467328

Define the causes the charging of an object?

Solution:

When an object looses or gains electrons by friction/conduction/induction, then it is charged.
Q 3251291124

What does the additive nature of electric charge mean?

Solution:

It means an electric charge is a scalar quantity and is added like algebraic numbers.
Q 3211467329

Define the cause of quantisation of electric charge?

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.
Q 3211567420

Explain the conservative nature of the electric force?

Solution:

The electric force is conservative in nature, because the work done by it in moving a charge is path independent.
Q 3221567421

Define the limitation of Coulomb's law?

Solution:

It is applied only for point charges.
Q 3231567422

Explain the epsi (absolute permittivity) signify?

Solution:

It is a measure of the degree to which a medium can resist the movement of charges.
Q 3241567423

What is the 1 coulomb (1 C) of electric charge.

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.
Q 3251567424

Write two properties of an electrostatic force.

Solution:

(a) It is conservative in nature.
(b) It depends on medium between the two charges.
Q 3261567425

How does the coulomb force between two point charges depend upon the dielectric constant of the intervening· medium?

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.
Q 3221391221

Name the physical quantity whose SI unit is V.m. Is it a vector or a scalar quantity?

Solution:

The physical quantity is an electric flux. It is a scalar quantity.
Q 3211567429

What is the dielectric constant of a medium. What is its unit?
CBSE 12 Delhi 2011
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.
Q 3261391225

When does a charged ring behave as a point charge?

Solution:

When the radius of ring is much smaller than the distance under consideration.
Q 3221667521

Explain the principle of superposition of forces.

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.
Q 3241667523

Draw the electric field lines due to a point charge (i) Q > 0 and (ii) Q < 0.

Solution:

Q 3211491320

Why do the electric field lines not form any dosed loops?

Solution:

Because they originate from positive ( +ve) charge and terminate at negative (-ve) charge.
Q 3221491321

Draw electric field lines for a system of two charges q1 and q2 such that
(i) q_1q_2 > 0; q_1 > q_2 > 0 (ii) q_1 q_2 < 0; q_1 > |-q_2| < 0, | q_1 | > |-q_2 |

Solution:

Q 3251667524

Define the physical significance of electric field?

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.
Q 3271667526

What is the term electric flux. Write its SI unit.

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
Q 3281667527

What is the term electric dipole moment. Is it a scalar or a vector quantity?

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.

### Concept Based Problems -Set 2

Set 2 - 1 , 2, 3 marks Problem
Q 3201667528

Define the an ideal (point) dipole?

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.
Q 3211491329

What is the direction of net force on electric dipole, placed in a non-uniform electric field?

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
Q 3211591420

When does an electric dipole placed in a non-uniform electric field experience a zero torque but non-zero force?

Solution:

When the dipole axis is parallel to the direction of electric field.
Q 3211767620

Define the Gaussian surface?

Solution:

A Gaussian surface is an imaginary closed surface in three dimensional space through with the flux of a vector field is calculated.
Q 3221767621

Explain the use of a Gaussian surface?

Solution:

A Gaussian surface is used to determine the electric field intensity around a point charge or charged body.
Q 3231767622

Why can a Gaussian. surface not pass through any discrete charge?

Solution:

Because the electric Geld due to a system of discrete charges is not defined at the location of any charge.
Q 3241791623

Sketch the electric field lines for the following system of charges.

Solution:

Q 3251791624

Explain the electric field intensity; Write its SI unit. Write the magnitude and direction of electric field intensity due to an electric dipole of length 2a at the mid-point of the line joining the two charges.

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).
Q 3271767626

Show that the electric field at the surface of a charged conductor is given by vecE = sigma/epsi_0 hatn where sigma is the surface charge density and ii is a unit vector normal to the surface in the outward direction.
CBSE-12-All-India 2010
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
Q 3261791625

A thin straight infinitely long conducting wire having charge density lamda is enclosed by a cylindrical surface of radius r and length l, its axis coinciding with the length of the wire. Find the expression for the electric flux through the surface of the cylinder.

CBSE-12-All-India 2011
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
Q 3261167925

Explain Gauss's theorem in electrostatics. Prove that no electric field exists inside a hollow charged sphe:re.

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
Q 3211178020

(a) What is the electric flux. Write its SI units.
(b) Using Gauss's law, prove that the electric field at a point due to a uniformly charged infinite plane sheet is independent of the distance from it.
(c) How is the field directed if (i) the sheet is positively charged, (ii) negatively charged?
CBSE 12 Foreign 2012
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.
Q 3231178022

Explain Gauss's law. Use it to deduce the expression for the electric field. due to a uniformly charged thin spherical shell at points

(i) inside and (ii) outside the shell.

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]