Chapter 1 Electric Charges And Fields Solutions
Question - 21 : - A conducting sphere ofradius 10 cm has an unknown charge. If the electric field 20 cm from the centreof the sphere is 1.5 × 103 N/Cand points radially inward, what is the net charge on the sphere?
Answer - 21 : -
Electric field intensity (E) at a distance (d)from the centre of a sphere containing net charge q isgiven by the relation,
Where,
q = Net charge = 1.5 × 103 N/C
d = Distance from the centre = 20 cm = 0.2 m
∈0 = Permittivity offree space
And,
= 9 × 109 N m2 C−2
∴
= 6.67 × 109 C
= 6.67 nC
Therefore, the net chargeon the sphere is 6.67 nC.
A uniformly chargedconducting sphere of 2.4 m diameter has a surface charge density of 80.0 μC/m2. (a) Find the charge on the sphere. (b)What is the total electric flux leaving the surface of the sphere?
Answer - 22 : -
(a) Diameter of the sphere, d = 2.4 m
Radius of the sphere, r = 1.2 m
Surfacecharge density, 
= 80.0 μC/m2 = 80 × 10−6 C/m2Total charge on thesurface of the sphere,
Q = Charge density × Surface area
=
= 80 × 10−6 × 4 × 3.14 × (1.2)2
= 1.447 × 10−3 C
Therefore, the charge onthe sphere is 1.447 × 10−3 C.
(b) Total electric flux (
) leaving out the surface of asphere containing net charge Q isgiven by the relation,
Where,
∈0 = Permittivity offree space
= 8.854 × 10−12 N−1C2 m−2
Q = 1.447 × 10−3 C
= 1.63 × 108 N C−1 m2
Therefore,the total electric flux leaving the surface of the sphere is 1.63 × 108 N C−1 m2.
An infinite line chargeproduces a field of 9 × 104 N/Cat a distance of 2 cm. Calculate the linear charge density.
Answer - 23 : -
Electric field produced bythe infinite line charges at a distance d havinglinear charge density λ isgiven by the relation,
Where,
d = 2 cm = 0.02 m
E = 9 × 104 N/C
∈0 = Permittivity offree space
= 9 × 109 N m2 C−2
= 10 μC/m
Therefore, the linearcharge density is 10 μC/m.
Question - 24 : - Two large, thin metal platesare parallel and close to each other. On their inner faces, the plates havesurface charge densities of opposite signs and of magnitude 17.0 × 10−22 C/m2.What is E: (a) in the outerregion of the first plate, (b) in the outer region of the second plate, and (c)between the plates?
Answer - 24 : -
The situation isrepresented in the following figure.
A and B are two parallelplates close to each other. Outer region of plate A is labelled as I, outer region of plate B is labelledas III, and the regionbetween the plates, A and B, is labelled as II.
Chargedensity of plate A,σ = 17.0 × 10−22 C/m2
Charge density of plate B, σ = −17.0 × 10−22 C/m2
In the regions, I and III, electric field E is zero. This is because charge isnot enclosed by the respective plates.
Electric field E in region II is given by the relation,
Where,
∈0 = Permittivity offree space = 8.854 × 10−12 N−1C2 m−2
∴
= 1.92 × 10−10 N/C
Therefore, electric fieldbetween the plates is 1.92 × 10−10 N/C.
Question - 25 : - An oil drop of 12 excesselectrons is held stationary under a constant electric field of 2.55 × 104 N C−1 inMillikan’s oil drop experiment. The density of the oil is 1.26 g cm−3. Estimate the radius of the drop. (g = 9.81 m s−2; e = 1.60 × 10−19 C).
Answer - 25 : -
Excess electrons on an oildrop, n = 12
Electric field intensity, E = 2.55 × 104 NC−1
Density of oil, ρ = 1.26 gm/cm3 =1.26 × 103 kg/m3
Acceleration due togravity, g = 9.81 m s−2
Charge on an electron, e = 1.6 × 10−19 C
Radius of the oil drop = r
Force (F) due to electric field E is equal to the weight of the oildrop (W)
F = W
Eq = mg
Ene 
Where,
q = Net charge on the oil drop = ne
m = Mass of the oil drop
= Volume of the oil drop ×Density of oil
= 9.82 × 10−4 mm
Therefore,the radius of the oil drop is 9.82 × 10−4 mm.
Which among the curvesshown in Fig. 1.35 cannot possibly represent electrostatic field lines?
(a) 
(b)
(c)
(d)
(e)
Answer - 26 : -
(a) The field lines showed in(a) do not represent electrostatic field lines because field lines must benormal to the surface of the conductor.
(b) The field lines showed in(b) do not represent electrostatic field lines because the field lines cannotemerge from a negative charge and cannot terminate at a positive charge.
(c) The field lines showed in(c) represent electrostatic field lines. This is because the field lines emergefrom the positive charges and repel each other.
(d) The field lines showed in(d) do not represent electrostatic field lines because the field lines shouldnot intersect each other.
(e) The field lines showed in(e) do not represent electrostatic field lines because closed loops are notformed in the area between the field lines.
Question - 27 : - In a certain region ofspace, electric field is along the z-direction throughout. The magnitude ofelectric field is, however, not constant but increases uniformly along thepositive z-direction, at the rate of 105 NC−1 permetre. What are the force and torque experienced by a system having a totaldipole moment equal to 10−7 Cm in the negative z-direction?
Answer - 27 : -
Dipole moment of thesystem, p = q × dl = −10−7 Cm
Rate of increase ofelectric field per unit length,
Force (F)experienced by the system is given by the relation,
F = qE
= −10−7 ×10−5
= −10−2 N
The force is −10−2 Nin the negative z-direction i.e., opposite to the direction of electric field.Hence, the angle between electric field and dipole moment is 180°.
Torque (τ) is givenby the relation,
τ = pE sin180°
= 0
Therefore, the torqueexperienced by the system is zero.
Question - 28 : - (a) A conductor A with acavity as shown in Fig. 1.36(a) is given a charge Q. Show that theentire charge must appear on the outer surface of the conductor. (b) Anotherconductor B with charge q is inserted into the cavity keepingB insulated from A. Show that the total charge on the outside surface of Ais Q + q [Fig. 1.36(b)]. (c) A sensitiveinstrument is to be shielded from the strong electrostatic fields in itsenvironment. Suggest a possible way.
Answer - 28 : -
(a) Let us consider a Gaussiansurface that is lying wholly within a conductor and enclosing the cavity. Theelectric field intensity E inside the charged conductor iszero.
Let q isthe charge inside the conductor and 
is the permittivity of freespace.
According to Gauss’s law,
Flux, 
Here, E =0
Therefore, charge insidethe conductor is zero.
The entire charge Q appearson the outer surface of the conductor.
(b) The outer surface ofconductor A has a charge of amount Q. Another conductor B havingcharge +q is kept inside conductor A and it is insulated from A.Hence, a charge of amount −q will be induced in the inner surfaceof conductor A and +q is induced on the outer surface of conductorA. Therefore, total charge on the outer surface of conductor A is Q + q.
(c) A sensitive instrument canbe shielded from the strong electrostatic field in its environment by enclosingit fully inside a metallic surface. A closed metallic body acts as anelectrostatic shield.
, where is the unit vector in the outwardnormal direction, and is the surface chargedensity near the hole.
Answer - 29 : -
Let us consider aconductor with a cavity or a hole. Electric field inside the cavity is zero.
Let E isthe electric field just outside the conductor, q is theelectric charge, is the charge density, and is the permittivity offree space.Charge 
According to Gauss’s law,
Therefore, the electric field just outside theconductor is
. This field is a superposition of field due to thecavity 
and the field due to the rest of the charged conductor. These fields are equaland opposite inside the conductor, and equal in magnitude and direction outsidethe conductor. Therefore, the field due to the rest of the conductoris.
Question - 30 : - Obtain the formula for theelectric field due to a long thin wire of uniform linear charge density λ withoutusing Gauss’s law. [Hint: Use Coulomb’s law directly and evaluatethe necessary integral.]
Answer - 30 : -
Take a long thin wire XY(as shown in the figure) of uniform linear charge density
Consider a point A at aperpendicular distance l from the mid-point O of the wire, asshown in the following figure.
Let E bethe electric field at point A due to the wire, XY.
Consider a small lengthelement dx on the wire section with OZ = x
Let q bethe charge on this piece.
Electric field due to thepiece,
The electric field isresolved into two rectangular components. 
is the perpendicularcomponent and 
is the parallel component.
When the whole wire isconsidered, the component is cancelled.
Only the perpendicularcomponent affects point A.
Hence,effective electric field at point A due to the element dx is dE1.
On differentiating equation (2), we obtain
From equation (2),
Putting equations (3) and(4) in equation (1), we obtain
The wire is so long that 
tends from 
to 
By integrating equation(5), we obtain the value of field E1 as,
Therefore, the electric field due to long wire is 