Video Lecture
Theory For Making Notes
Electric Field
Electric Field is the space around a charge in which its influence (electric forces) can be experienced. In principle the electric field of a charge extends upto infinity.
In other words an electric charge influences its surroundings by creating an electric field. The second charge does not interact directly with the first, rather, it responds to whatever field it encounters.
Electric Field Intensity
The electric field strength or intensity due to a point source charge q at any point P at a distance r from it is given by the force acting on a unit positive test charge placed at that point.
The electric field strength E at a point is defined as the force F per unit charge experienced by a vanishing small positive test charge q0

E=q0→0limq0F
E is also called the electric field intensity or simply the electric field.
Note that E at a point is equal to the force experienced by a unit positive charge placed at that point. Obviously, if E is the electric field at any point, then the force experienced by a charge q at that point is
F= qE
SI unit of E is Newton per coulomb (N/C), which is same as volt per metre (V/m):
unit of E = CN = mJ × C1 = CJ × m1 = mV
The dimensions of E are [MLT–3A–1].
The electric field created by a point charge q is given by
E = r2kqr^
where the unit vector r has it origin at the source charge q as shown in the figure

Vector form of electric field at position vector r due to a point charge q placed at a point whose position vector is r0 is given as E=∣r−r0∣3kq(r−r0)
The electric field due to a charge points away from the charge if it is positive and towards it if the charge is negative as shown in the diagram given below


Principle of superposition
Since the principle of superposition is valid for Coulomb’s law, it is also valid for the electric field. To calculate the field strength at point due to a system of charges, we first find the individual field intensity E1 due to q1, E2 due to q2 and so on.
For N point charges, the resultant field intensity is the vector sum of all the individual electric fields due to different charges, so if E is the net electric field then it is given by
E = E1 +E2 + …….. + EN

Let the point charges q1,q2,q3,…… placed at the points whose position vectors are r1,r2,r3,……. respectively. Then their net electric field at a point whose position vector r is given as
E=∣r−r1∣3kq1(r−r1)+∣r−r2∣3kq2(r−r2)+∣r−r3∣3kq3(r−r3)+……..=ki=1∑i=N∣r−ri∣3kqi(r−ri)
This is also known as superposition principle.
Motion of a charged particle in an electric field
Let a particle of mass m, charge q be situated in an electric field of strength E, then force on the charged particle is given by F→=q.E→
The acceleration a produced by this force is a→=mF→=mqE→ …(i)
If the particle is at rest at t = 0, then velocity of charged particle after time t is
v=u+at=0+mqEt
or v=mqEt …(ii)
and distance traversed by the particle is s=ut+21(mqE)t2
or s=21(mqE)t2 …(iii)
and kinetic energy K=21m(mqEt2) =21mm2q2E2t2
K=2mq2E2t2 …(iv)
If a charged particle enter in field in perpendicular direction, then force on the particle is along Y−axis is given by F→Y=q.E→

Acceleration of the particle along Y–axis a→Y=mF→Y
Initial velocity is u along x – axis and it is zero along Y–axis. The deflection after time t along y-axis is
y=21mqEt2 …(v)
As there is no acceleration along X-axis, therefore the distance traversed by particle in time t along X–axis
x=ut …(vi)
From (v) and (vi), we get
y=21.mqE(ux)2⇒y∝x2
This shows that the path of the charged particle in perpendicular field is a parabola.
Electric field due to a uniformly charged ring on its axis.
Let a total positive charge Q is distributed uniformly around a thin, circular, non-conducting ring of radius a. Let us calculate its electric field E at a point P along the axis of the ring, at a distance x from the centre, as shown in figure.
Consider a element charge dq on the ring and its electric field is written at the point P
As E=r2kdqr^ , where the magnitude dE=(a2+x2)kdq . We can find its two components one along the x axis i.e. dEx the other perpendicular to x axis i.e. dE⊥
Note the symmetry. Every element dq can be paired with a similar element on the opposite side of the ring. Every component dE⊥ perpendicular to the x-axis is thus cancelled by an equal component dE⊥ in the opposite direction. In the summation process, all the perpendicular components dE⊥ add to zero. Thus, we are to add only the dE⊥ components.

Its x-component is
dEx=dEcosθ=(a2+x2)kdq.a2+x2x
E = Ex = ∫dEx=k∫(a2+x2)3/2xdq
As we integrate around the ring,
∫dq=Q
E=(a2+x2)3/2kxQ
Note that At centre x = 0 so Ecentre = 0
Whereas at a point on the axis such that x>>a. On neglecting a from the denominator we get E=x2kQ
If we differentiate E with respect to x and putting dE/dx equal to zero we get x=±2R, It is the point where the electric field becomes maximum with magnitude Emax=63πε0a2Q
Graph showing the variation of electric field of a ring with distance x

Electric field due to charged Disc On its Axis
Let us calculate the field intensity at a distance y from the centre along the central axis of a circular disc of radius R and surface charge density σ(C/m2).
Due to circular symmetry of the disc, a ring of radius x and width dx is chosen. All points of the ring are equidistant from the point P.
Consider a component of the field parallel to the plane of the disc. Any contribution of this component from some point on the ring has an equal and opposite contribution from the diametrically opposite part of the ring.
Thus, due to symmetry, there is no component of the field parallel to the plane of the disc
i.e. Ex= 0
The y-component of the field is
dEy = dE cos θ = r2kdqry

where dq = σdA = σ(2pxdx)
and r2 = x2 + y2
dEy= (x2+y2)3/2kσ(2πxdx)y

On integrating
Ey= πkσy0∫R(x2+y2)3/22xdx
or Ey= 2πkσ[1−(R2+y2)1/2y] (i)
Note
When y >>R, using binomial theorem.
(y2+R2)1/2y=(1+y2R2)−1/2≈1−2y2R2
Thus, the expression reduces to
Ey= y2πkσR2
or Ey= y2kq since q = s(pR2)
Which represents the electric field due to a point charge
As R →∞, the disc behaves like an infinite sheet and the expression (i) reduces to
E = 2πkσ = σ/2εo
Illustration
A 5.0µC point charge is placed at the point x = 0.2 cm, y = 0.3 cm. Find the magnitude of E due to it (a) at the origin and (b) at point A with coordinates x = 1.0m, y = 1.0 m.

Solution
(a) The direction of E at the origin is shown in figure
The distance of point O from the charge is
r=(0.2)2+(0.3)2=0.36m
also tanθ=0.20.3=1.5 Þ q = tan–1 1.5 = 56.3°
The magnitude of electric intensity is
E=r2kq=(0.36)2(9×109)(5×10−6)=3.46×105NC−1

(b) The distance between the charge q and the point A is
r=(1.0−0.2)2+(1.0−0.3)2=1.06m
tanθ=0.80.7=87
θ=tan−187=41.2∘
The magnitude of E is
E=r2kq=(1.06)2(9×109)(5×10−6)=4×104N
Illustration
The figure shows a system of two charges. A point charge q1 = 20nC is at (-1,0) while q2 = -10nC is at (+1,0). Find the resultant field intensity at a point P with coordinates (2,2).

Solution
1.
The field vector E1 points away from the charge q1 while the vector E2 points toward the charge q2.
2.
The distance r1 and r2 are
r1 = (2+1)2+22=13=3.6m
r2 = 12+22=5=2.2m
The magnitude of the fields are
E1 = r12kq1=(3.6)2(9×109)(20×10−9) =13.8NC−1
E2 = r22kq2=(2.2)2(9×109)(10×10−9) =18NC−1
3.
The components of the resultant field are

Ex=E1cosθ1−E2cosθ2
Ey=E1cosθ1−E2cosθ2
were cosθ1 =3.63 ; cosθ2= 2.21
and sinθ1 = 3.62 ; sinθ2 = 2.22
Thus, Ex = (13.8) 3.63−18(2.21)=3.3N/C
Ey = (13.8) 3.62 – 18 (2.22) = -8.7 NC–1
Finally, E=3.3i^−8.7j^NC−1
Illustration
Three point charges are placed at the corners of an equilateral triangle as shown in figure. What is the value of electric field at the centre.

Solution
Let us take origin at the centre and x and y-axis as shown in figure.
E1, E2 and E3 are the electric field due to +Q, +Q and – Q.
E1=4π∈0(3a)2Q(cos30∘i^+sin30∘j^)
E2=4π∈0(3a)2Q(−cos30∘i^+sin30∘j^)
E3=4π∈0(3a)2Q(+j^)
Enet=E1+E2+E3
=4π∈0a26Qj^
Illustration
Find the field intensity at a point P at a distance R from a finite line of charge of length L and linear charge density l C/m, as shown in the Fig. (A).

Solution
Consider a small element of length dy at a distance y, as shown in Fig. (B). The magnitude of the contribution to the field at point P due to this element is
dE=r2k λdy …(i)
To carry out the integration we express the variables in terms of the angle q. From the figure,
r = R sec q and y = R tan q
on differentiating, dy = R sec2q. Thus, Eqn. (i) becomes
dE=(Rsec θ)2k λ(Rsec2θ dθ)=Rk λ dθ
The components of dE along x-axis and y-axis are
dEx = dE cos q; dEy = dE sin q
On integrating, we get
Ex=Rk λ−β∫+αcos θ dθ =Rk λ (sin θ)∣−β+α = Rk λ (sin α + sin β)
and Ey=Rk λ−β∫+αsin θ dθ =Rk λ (−cos θ)∣−β+α = Rkλ (cos β − cos α)
The magnitude of the net field at point P will be
E=Ex2+Ey2
Important Points
(i)
If the point P is at the perpendicular bisector of the line charge then a = b
andEx = Rk λ 2 sin a = Rk λ (L╱2)2+R2L
Ey = 0
(ii)
If the charge were distributed over an infinite length the both a and b approaches π/2.
Illustration
A thin conducting ring of radius R is charged with a linear densityλ=λ0cosφ , where λ0 is a positive constant and φ is the angle shown in fig. Find the electric field intensity E at the center of the ring.

Solution
Consider a differential element of charge dq whose field at center O.

the components of its field are dEx=dEcosφ=R2kdqcosφ=R2kλRφcosφ
Ex=∫dEx=Rkλ00∫2πcos2φdφ
=2Rkλ00∫2π(1−cos2φ)dφ
or Ex=2Rkλ0[0∫2πdφ−0∫2πcos2φdφ]
=2Rkλ02π=Rkλ0π=4ε0Rλ0
Similarly,
Ey=0∫2π2Rkλ0sinφcosφdφ
or Ey=2Rkλ00∫2πsin2φdφ=4Rkλ0[−cos2φ]02π=0
Hence the net field at O is along the x−axis and its magnitude is 4ε0Rλ0. The symmetry of this distribution also implies that the field is directed along x−axis.
Practice Questions (Basic Level)
Q.1
Figure shows a charge +Q a distance 2d from a charge –Q and a point X distance d from –Q. The field strength at X is numerically, using SI units

(a) 4πεod2Q
(b) 36πεod2Q
(c) 4πεod23Q
(d) 9πεod22Q
Ans .(d)
Q.2
A positively charged pendulum is oscillating in a uniform electric field as shown in the figure. Its time period as compared to that when it was uncharged

(a) will increase
(b) will decrease
(c) will not change
(d) will first increase then decrease
Ans. (a)
Q.3
A drop of oil density r and radius r carries a charge q when placed in a electric field E, its moves upwards with a velocity n.
If r0 is the density of air, h be the viscosity of the air, then which of the following forces is directed upwards ?
(a) 6 p h r n
(b) q E
(c) 34πr3(ρ−ρ0)g
(d) qE + 6 p r h n
Ans. (b)
Q.4
The diagram shows electric lines of force. If EA and EB are electric fields at A and B and distance AB = r, then

(a) EA > EB
(b) EA = EB/r
(c) EA < EB
(d EA = EB/r
Ans (a)
Q.5
Charges 2Q and –Q are placed as shown. The point at which electric field intensity is zero will be

(a) somewhere between –Q and 2Q
(b) somewhere on the left of –Q
(c) somewhere on the right of 2Q
(d) somewhere on the right bisector of line joining –Q and 2Q
Ans. (b)
Q.6
Five point charges (+q each) are placed at the five vertices of a regular hexagon of side 2a. What is the magnitude of the net electric field at the centre of the hexagon?
(a) 4πεo1a2q
(b) 16πεoa2q
(c) 4πεoa22q
(d) 16πεoa25q
Ans . (b)
Q. 7
In a regular polygon of n sides, each corner is at a distance r from the centre. Identical charges of magnitude Q are placed at (n-1) corners. The field at the centre is
(a) k r2Q
(b) (n-1)k r2Q
(c) n−1nk r2Q
(d) nn−1k r2Q
Ans . (a)
Q.8
Charge Q is given a displacement =aî+bĵ in an electric field =E1î+E2ĵ. The work done is
(a) Q(E1a+E2b)
(b) Q (E1a)2+(E2b)2
(c) Q(E1+E2) a2+b2
(d) Q (E12+(E22
Ans .(a)
Q.9
Which of the following is not true for a region with a uniform electric field?
(a) It can have free charges
(b) It may have uniformly distributed charge.
(c) It may contain dipoles.
(d) None of the above.
Ans. (d)
Q.10
A simple pendulum of time period T is suspended above a large horizontal metal sheet with uniformly distributed positive charge. If the bob is given some negative charge, its time period of oscillation will be
(a) >T
(b) <T
(c) T
(d) proportional to its amplitude.
Ans. (b)
Q. 11
The maximum electric field intensity on the axis of a uniformly charged ring of charge q and radius R will be :
(a) 4πε0133R2q
(b) 4πε013R22q
(c) 4πε0133R22q
(d) 4πε0122R23q
Ans. (c)
Q.12
Two particles A and B (B is right of A) having charges and , respectively, are held fixed with separation of 20 cm. Where should a third charged particle be placed so that it does not experience a net electric force.
(a) 5cm right of B
(b) 5cm left of A
(c) 20cm left of A
(d) 20cm right of B
Ans. (d)
Q.13
Two particles of masses in the ratio 1 : 2, with charges in the ratio 1 : 1, are placed at rest in a uniform electric field. They are released and allowed to move for the same time. The ratio of their kinetic energies will be finally
(a) 2 : 1
(b) 8 : 1
(c) 4 : 1
(d) 1 : 4
Ans. (a)
Q.14
An electron of mass , initially at rest, moves through a certain distance in a uniform electric field in time . A proton of mass also initially at rest, takes time to move through an equal distance in this uniform electric field. Neglecting the effect of gravity, the ratio is nearly equal to
(a) 1
(b) Mp/Me
(c) Me/Mp
(d) 1836
Ans. (b)
15.
Two point charges +8q and -2q are located at x = 0 and x = L respectively. The location of a point on the x-axis at which the net electric field due to these two point charges is zero, is
(a) 2 L
(b) 4L
(c) 8 L
(d) 4 L
Ans (a)
Practice Questions (JEE Main Level)
Q.1
The intensity of electric field due to a ring of radius R at a distance x from its centre on the axis of ring carrying charge Q is given by
E = 4πεo1(R2+x2)3/2Qx
The distance from the centre of the ring where intensity of electric field will be maximum is
(a) x = R
(b) x = R/2
(c) x = 2R
(d) x = 2R
Ans. (d)
Q.2
A spring-block system undergoes vertical oscillation above a large horizontal metal sheet with uniform positive charge. The time period of the oscillation is T. If the block is given a charge Q, its time period of oscillation will be
(a) T (b) >T (c) T if Q is positive and
Ans. (a)
Q.3
A large flat metal surface has a uniform charge density +s. An electron of mass m and charge e leaves the surface at point A with speed u, and returns to it at point B. Disregard gravity. The maximum value of AB is
(a) σeu2mε0
(b) mσu2eε0
(c) ε0σmu2e
(d) ε0mu2σe
Ans. (a)
Q.4
An extremely long wire is uniformly charged. An electron is revolving about the wire and making 108 revolutions per second in an orbit of radius 2cm. Linear charge density of the wire is nearly :
(a) 50 nC (b) 25 nC (c) 62.5 nC (d) 12.5 nC
Ans. (a)
Q.5
Four identical charges Q are fixed at the four corners of a square of side a. The electric field at a point P located symmetrically at a distance from the center of the square is
(a) 22πε0a2Q
(b) 2πε0a2Q
(c) πε0a222Q
(d) πε0a22Q
Ans. (b)
Q.6
One of the two parallel metallic plates is uniformly charged with charge +q and the other one is charged with charge –q. In this case, the electric field between them is E. When the negatively charged plate is discharged then recharged with a positive charge 4q, the magnitude of electric field between the plates becomes
(a) 1.5 E (b) 2.5 E (c) 3 E (d) 5 E
Ans..(a)
Q.7
Consider three identical metal spheres A, B and C. Spheres A carries charge + 6q and sphere B carries charge – 3 Sphere C carries no charge. Spheres A and B are touched together and then separated. Sphere C is then touched to sphere A and separated from it. Finally the sphere C is touched to sphere B and separated from it. Find the final charge on the sphere C.
Ans. 89Q
Q.8
Two positive point charges of magnitude Q each are separated by a distance “2a”. A test charge is located in a plane which is normal to the line joining these charges and midway between them. The locus of the points in this plane for which the force on the test charge has a maximum value is
(a) a circle of radius r=a/2
(b) a circle of radius r=a
(c) a straight line
(d) an ellipse
Ans. (a)
Q.9
A particle of mass m and charge q is placed at rest in a uniform electric field E as shown and released. The kinetic energy it attains after moving a distance y is

(a) 21qEy
(b) qEy
(c) qE2y
(d) 21m(qEy)
Ans. (b)
Q.10
The electric field on two sides of a charged plate is shown in the figure. The charge density on the plate is given by

(a) 2eo
(b) 4eo
(c) 10eo
(d) zero
Ans. (b)
Q.11
A thin glass rod is bent into a semicircle of radius r. A charge +Q is uniformly distributed along the upper half, and a charge –Q is uniformly distributed along the lower half. The electric field E at P, the center of the semicircle, is

(a) π2ε0r2Q
(b) π2ε0r22Q
(c) π2ε0r24Q
(d) 4π2ε0r2Q
Ans. (a)
Q.12
A block of mass m having charge q, is hinged by a spring of spring constant k in a vertical electrostatic field E. The spring extension in equilibrium will be

(A)kmg
(B) kqE
(C) kmg+qE
(D) kmg−qE
Q. 13
Two identical point charges are placed at a separation of l. P is a point on the line joining the charges, at a distance x from any one charge. The field at P is E. E is plotted against x for values of x from close to zero to slightly less than l. Which of the following best represents the resulting curve?

Ans. (d)
Q.14
Figure shows a uniformly charged hemispherical shell. The direction of electric field at point p, that is off-centre (but in the plane of the largest circle of the hemisphere), will be along

(a) pa
(b) pb
(c) pc
(d) pd
Ans. (b)
Q.15
Three charges + 2Q, – 3Q and –Q are arranged in a line as shown in the figure. The number of points along the line at which the net electric field is zero is

(a) 1
(b) 2
(c) 3
(d) More than 3
Ans. (b)
Practice Questions (JEE Advance Level)
Q.1
A particle of mass m and charge –q is projected from the origin with a horizontal speed v into an electric field of intensity E directed downward. Choose the wrong statement.

(a) The kinetic energy after a displacement y is qEy.
(b) The horizontal and vertical components of acceleration are ax = 0, ay = mqE
(c) The equation of trajectory is y = 21(mv2qEx2)
(d) The horizontal and vertical displacements x and y after a time t are x = vt and
y = 21ayt2
Ans. (a)
Q.2
A small ball of mass m and charge +q tied with a string of length l, is rotating in a vertical circle under gravity and a uniform horizontal electric field E as shown. The tension in the string will be minimum for

(a) θ=tan−1(mgqE)
(b) q = p
(c) q = p – tan−1(mgqE)
(d) q = p + tan−1(mgqE)
Ans. (d)
Q.3
The variation of electric field between the two charges q1 and q2 along the line joining the charges is plotted against distance from q1 (taking rightward direction of electric field as positive) as shown in the figure. Then the correct statement is

(a) q1 and q2 are positive and q1 < q2
(b) q1 and q2 are positive and q1 > q2
(c) q1 is positive and q2 is negative and q1 < q2
(d) q1 and q2 are negative and q1 < q2
Ans. (a)
Q.4
Consider a regular cube with positive point charge + Q in all corners except for one which has a negative point charge – Q. Let the distance from any corner to the center of the cube be r. The magnitude of electric field at point P, the center of the cube will be

(a) E = 7keQ/r2
(b) E = 1keQ/r2
(c) E = 2keQ/r2
(d) E = 6keQ/r2
Ans. (d)
Q.5
The figure shows, two point charges q1 = 2Q ( > 0) and
q2 = –Q. The charges divide the line joining them in three parts I, II and III.

(a) Region III has a local maxima of electric field.
(b) Region I has a local minima of electric field.
(c) Equilibrium position for a test charge lies in region II.
(d) The equilibrium for constrained motion along the line joining the charges is stable for a positive test charge.
Ans. (a)
Q.6
A simple pendulum of length l and bob mass m is hanging in front of a large nonconducting sheet having surface charge density σ=q2mg∈0 . If suddenly a charge +q is given to the bob and it is released from the position shown in figure. The maximum angle through which the string is deflected from vertical will be

(a) 90°
(b) 60°
(c) 45°
(d) 30°
Ans. (a)
Q.7
A particle of mass m and negative charge q is thrown in a gravity free space with speed u from the point A on the large non conducting charged sheet with surface charge density s, as shown in figure. The maximum distance from A on sheet where the particle can strike will be

(a) qσmu2ε0
(b) qσ2 mu2ε0
(c) qσ2mu2ε0
(d) 2qσmu2ε0
Ans. (c)
Q.8
A square loop of side ‘l‘ each side having uniform linear charge density ‘l’ is placed in ‘xy’ plane as shown in the figure. There exists a non-uniform electric field E=la(x+l)i^ where a and l are constants and x is the position of the point from origin along x-axis. Find the resultant electric force on the loop (in Newtons), if l = 10 cm, l = 20 mC/m and a = 5 × 105 N/C.

(a) 9N
(b) 6N
(c) 45N
(d) 30N
Ans. (b)
Q. 9
A positively charged particle of charge q and mass m is suspended from a point by a string of length l. In the space a uniform horizontal electric field E exists. The particle is drawn aside so that the string becomes vertical and then it is projected horizontally with velocity v such that the particle starts to move along a circle with same constant speed v. Find the speed v.

(a) 9N
(b) 6N
(c) 45N
(d) 30N
Ans. (b)
Q. 9
A positively charged particle of charge q and mass m is suspended from a point by a string of length l. In the space a uniform horizontal electric field E exists. The particle is drawn aside so that the string becomes vertical and then it is projected horizontally with velocity v such that the particle starts to move along a circle with same constant speed v. Find the speed v.

(a) πε0mReλ(463−22)
(b) πε0mReλ(463+22)
(c) πε0mReλ(433+22)
(d) None of these
Ans. (a)
Q.11
The charge per unit length of the four quadrant of the ring is 2l, – 2l, l and – l respectively. The electric field at the centre is

(a) 2πε0Rλ i^
(b) 2πε0Rλ j^
(c) 4πε0R2 λ i^
(d) None of these
Ans. (b)
Q.12
An inclined plane making an angle of 30° with the horizontal is placed in a uniform electric field E of 100 V/m (see figure). A particle of mass 1 kg and charge 0.01 C is allowed to slide down from rest from a height of 1 m. If the coefficient of friction is 0.2, the time taken by the particle to reach the bottom will be

(a) 1.2 sec
(b) 1.337 sec
(c) 1.447 sec
(d) 1.5 sec
Ans. (a)
Q.13
A particle of mass m and charge Q is placed in an electric field E which varies with time t as . It will undergo simple harmonic motion of amplitude.
(a) mω2QE02
(b) mω2QE0
(c) mωQE0
(d) mωQE0
Ans. (b)
Q.14
A nonconducting ring of radius R has uniformly distributed positive charge Q. A small part of the ring, of length d, is removed (d << R). The electric field at the centre of the ring will now be
(a) directed towards the gap, inversely proportional to R3.
(b) directed towards the gap, inversely proportional to R2.
(c) directed away from the gap, inversely proportional to R3.
(d) directed away from the gap, inversely proportional to R2.
Ans. (a)
Q.15
Electric field in a space is given as E=E0x(x−1)(x−2)(x−3) N/C along X. axis. Where E0 is a positive constant.
(a) Electric field is along positive direction at x=21
(b) Electric field is along negative direction at x=23
(c) A positive charge is in stable equilibrium at x=2
(d) A negative charge is in stable equilibrium at x=2
Ans. (c)
Q.16
Two infinitely long charged wires with linear densities l and 3l are placed along x and y axis respectively determined the slope of electric field and any point on the line y=3x.
Ans. 30°
Q.17
Three charges + 2Q, – 3Q and –Q are arranged in a line as shown in the figure. The number of points along the line at which the net electric field is zero is

(a) 1 (b) 2 (c) 3 (d) More than 3
Ans. (b)
18.
Two semi-infinite line charges with linear density +λ C/m and –λ C/m lie along the x-axis with their ends located at x = –a and x = +a, as shown in fig. What is the electric field strength at the origin ?

(a) a2kλ
(b) akλ
(c) a9kλ
(d) a5kλ
Ans (a)
19.
Three infinite parallel charged sheets are placed as shown .The electric field on the four regions indicated in the figure in respective order must be

(a) EI=5εoσ,EII=2εo5σ,EIII=2εoσ,EIV=−5εoσ
(b) EI=2εoσ,EII=2εo2σ,EIII=2εoσ,EIV=−2εoσ
(c) EI=2εoσ,EII=2εo5σ,EIII=2εoσ,EIV=−2εoσ
(d) EI=2εoσ,EII=2εo5σ,EIII=2εoσ,EIV=−2εoσ
Ans : (d)
20.
Two infinite lines of charge with equal linear charge densities λ C/m are placed along the x and y axes, as in figure. What is the electric field strength at an arbitrary point (x, y) ?

(a) 2πεoλ (x1i^+y1j^)
(b) piεoλ (x1i^+y1j^)
(c) 1πεoλ (x1i^+y1j^)
(d) 8πεoλ (x1i^+y1j^)
Ans (a)
Comprehension Question (21 To 23)
An electron is accelerated from rest by a uniform field of magnitude 105 N/C
(21)
How long does it take to reach 0.1c, where c = 3 ´ 10 8 m/s is the speed of light ?
(i) 1.5 ns
(ii) 1.3 ns
(iii) 1.7 ns
(iv)1.2 ns
Ans (iii)
(22)
How far does it travel in this time?
(i) 2.00 cm
(ii) 1.5 cm
(iii) 2.56 cm
(iv) 3.5 cm
Ans (iii)
(23)
What is the final kinetic energy?
Neglect relativistic effects.
(i) 4.1 ´ 10–16 J
(ii) 3.5 ´ 10–16 J
(iii) 2 ´ 10–16 J
(iv) 7 ´ 10–16 J
Ans (i)
24.
Point charge q and -q are located at the vertices of a square with diagonals 2l as shown in figure. Find the magnitude of the electric field strength at a point located symmetrically with respect to the vertices of the square at a distance x from its center.

(a) 2πεo(l2+x2)1/2ql
(b) 3πεo(l2+x2)4/6ql
(c) 2πεo(l5+x2)1/5ql
(d) 2πεo(l2+x2)3/2ql
Ans (d)
25.
An electron is fired with an initial speed vo at 45° to the horizontal from the bottom plate of a parallel plate arrangement as shown. The plates are separated by 2 cm and are very long. What is the maximum value of vofor the electron not to hit the upper plate ? Take E = 103 N/C.

(a) 1.75 ´ 106 m/s
(b) 3.75 ´ 106 m/s
(c) 5.75 ´ 106 m/s
(d) 3.5 ´ 106 m/s
Ans (b)