Video Lecture
Theory For Notes Making
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 \vec{E} at a point is defined as the force \vec{F} per unit charge experienced by a vanishing small positive test charge q0
\displaystyle \vec{E}=\underset{{{{q}_{0}}\to 0}}{\mathop{{\lim }}}\,\frac{F}{{{{q}_{0}}}}
\vec{E} is also called the electric field intensity or simply the electric field.
Note that \vec{E} at a point is equal to the force experienced by a unit positive charge placed at that point. Obviously, if \vec{E} is the electric field at any point, then the force experienced by a charge q at that point is
\vec{F}= q \vec{E}
SI unit of \displaystyle \vec{E} is Newton per coulomb (N/C), which is same as volt per metre (V/m):
unit of E\text{ }=\text{ }\frac{\text{N}}{\text{C}}\text{ }=\text{ }\frac{\text{J}}{\text{m}}\text{ }\times \text{ }\frac{\text{1}}{\text{C}}\text{ }=\text{ }\frac{\text{J}}{\text{C}}\text{ }\times \text{ }\frac{\text{1}}{\text{m}}\text{ }=\text{ }\frac{\text{V}}{\text{m}}
The dimensions of E are [MLT–3A–1].
The electric field created by a point charge q is given by
\displaystyle \vec{E} = \displaystyle \frac{{kq}}{{{{r}^{2}}}}\hat{r}
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 \displaystyle \vec{r} due to a point charge q placed at a point whose position vector is \displaystyle {{\vec{r}}_{0}} is given as \displaystyle \vec{E}=\frac{{kq\left( {\vec{r}-{{{\vec{r}}}_{0}}} \right)}}{{|\vec{r}-{{{\vec{r}}}_{0}}{{|}^{3}}}}
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 {{\vec{E}}_{1}} due to q1, {{\vec{E}}_{2}} 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 \vec{E} is the net electric field then it is given by
\vec{E} = {{\vec{E}}_{1}} + {{\vec{E}}_{2}} + …….. + {{\vec{E}}_{N}}
Let the point charges \displaystyle {{q}_{1}},\,\,{{q}_{2}},\,\,{{q}_{3}},\,\,…… placed at the points whose position vectors are \displaystyle {{\vec{r}}_{1}},\,\,{{\vec{r}}_{2}},\,\,{{\vec{r}}_{3}},\,\,……. respectively. Then their net electric field at a point whose position vector \displaystyle \vec{r} is given as
\displaystyle \vec{E}\,\,=\,\frac{{k{{q}_{1}}\,\left( {\vec{r}-{{{\vec{r}}}_{1}}} \right)}}{{|\vec{r}-{{{\vec{r}}}_{1}}{{|}^{3}}}}\,+\,\frac{{k{{q}_{2}}\,\left( {\vec{r}-{{{\vec{r}}}_{2}}} \right)}}{{|\vec{r}-{{{\vec{r}}}_{2}}{{|}^{3}}}}\,+\,\frac{{k{{q}_{3}}\,\left( {\vec{r}-{{{\vec{r}}}_{3}}} \right)}}{{|\vec{r}-{{{\vec{r}}}_{3}}{{|}^{3}}}}+\,……..
hence \displaystyle \vec{E}\,\,=k\sum\limits_{{i=1}}^{{i=N}}{{\frac{{k{{q}_{i}}\,\left( {\vec{r}-{{{\vec{r}}}_{i}}} \right)}}{{|\vec{r}-{{{\vec{r}}}_{i}}{{|}^{3}}}}}}
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 \overset{\to }{\mathop{F}}\,=q.\overset{\to }{\mathop{E}}\,
The acceleration a produced by this force is \overset{\to }{\mathop{a}}\,=\frac{{\overset{\to }{\mathop{F}}\,}}{m}=\frac{{q\overset{\to }{\mathop{E}}\,}}{m} …(i)
If the particle is at rest at t = 0, then velocity of charged particle after time t is
\vec{v}=\vec{u}+\vec{a}t=0+\frac{{q\vec{E}}}{m}t
or \vec{v}=\frac{{q\vec{E}}}{m}t …(ii)
and distance traversed by the particle is \vec{s}=\vec{u}t+\frac{1}{2}\left( {\frac{{q\vec{E}}}{m}} \right){{t}^{2}}
or \vec{s}=\frac{1}{2}\left( {\frac{{q\vec{E}}}{m}} \right){{t}^{2}} …(iii)
and kinetic energy K =\frac{1}{2}m\left( {\frac{{qE}}{m}{{t}^{2}}} \right) =\frac{1}{2}m\frac{{{{q}^{2}}{{E}^{2}}}}{{{{m}^{2}}}}{{t}^{2}}
K=\frac{{{{q}^{2}}{{E}^{2}}}}{{2m}}{{t}^{2}} …(iv)
Motion of a charged particle when its initial velocity is perpendicular to a uniform electric field :
Consider a charged particle given an initial velocity along +ve x axis in a uniform electric field acting along +ve y axis.
The force on the particle is along y-axis is given by {{\overset{\to }{\mathop{F}}\,}_{y}}=q.\overset{\to }{\mathop{E}}\,
Acceleration of the particle along y–axis must be given by a=\frac{F}{m}
hence {{a}_{y}}= \frac{{F}_{y}}{m}=\frac{{qE}}{m}
Initial velocity is u along x – axis and it is zero along y–axis. The deflection (or displacement) after time t along y-axis can be found using S=ut+\frac{1}{2}a{{t}^{2}}
hence y=\frac{1}{2}\frac{{qE}}{m}{{t}^{2}} …(v)
As there is no acceleration along x-axis, therefore the displacement of particle in time t along x–axis is
x=ut …(vi)
Eleminating t from equation (v) and (vi), we get equation of trajectory as
y=\frac{1}{2}.\frac{{qE}}{m}{{\left( {\frac{x}{u}} \right)}^{2}}
\Rightarrow y\propto {{x}^{2}}
This shows that the path of the charged particle 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 \displaystyle \vec{E}=\frac{{kdq}}{{{{r}^{2}}}}\hat{r} , where the magnitude \displaystyle dE = \frac{{k dq}}{{({{a}^{2}}+{{x}^{2}})}} . 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
\displaystyle d{{E}_{\text{x}}} = dE \text{cos} \theta = \frac{{k dq}}{{({{a}^{2}}+{{x}^{2}})}}.\frac{x}{{\sqrt{{{{a}^{2}}+{{x}^{2}}}}}}
E = Ex = \displaystyle \int{{d{{E}_{x}}}} = k\int{{\frac{{x dq}}{{{{{({{a}^{2}}+{{x}^{2}})}}^{{3/2}}}}}}}
As we integrate around the ring,
so \displaystyle \int{{dq}}=Q
Hence \displaystyle E = \frac{{k x Q}}{{{{{({{a}^{2}}+{{x}^{2}})}}^{{3/2}}}}}
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=\frac{{kQ}}{{{{x}^{2}}}}
NOTE
If we differentiate E with respect to x and putting dE/dx equal to zero we get x=\pm \frac{R}{{\sqrt{2}}}, It is the point where the electric field becomes maximum with magnitude {{E}_{{\max }}}=\frac{Q}{{6\sqrt{3}\pi {{\varepsilon }_{0}}{{a}^{2}}}}
Graph showing the variation of electric field of a ring with distance x
Illustration
A 5.0µC point charge is placed at the point x = 0.2 cm, y = 0.3 cm. Find the magnitude of \displaystyle \overrightarrow{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 \displaystyle \overrightarrow{E} at the origin is shown in figure
The distance of point O from the charge is
\displaystyle r=\sqrt{{{{{(0.2)}}^{2}}+{{{(0.3)}}^{2}}}}=0.36\text{m}
also \displaystyle \tan \theta =\frac{{0.3}}{{0.2}}=1.5
⇒ θ = tan–1 1.5 = 56.3°
The magnitude of electric intensity is
\displaystyle E=\frac{{kq}}{{{{r}^{2}}}}=\frac{{(9\times {{{10}}^{9}})\,\,(5\times {{{10}}^{{-6}}})}}{{{{{(0.36)}}^{2}}}}=3.46\times {{10}^{5}}\text{N}{{\text{C}}^{{-1}}}
(b) The distance between the charge q and the point A is
\displaystyle r=\sqrt{{{{{(1.0-0.2)}}^{2}}+{{{(1.0-0.3)}}^{2}}}}=1.06\text{m}
\displaystyle \tan \theta =\frac{{0.7}}{{0.8}}=\frac{7}{8}
\displaystyle \theta ={{\tan }^{{-1}}}\frac{7}{8}=41.2{}^\circ
The magnitude of E is
\displaystyle E=\frac{{kq}}{{{{r}^{2}}}}=\frac{{(9\times {{{10}}^{9}})\,(5\times {{{10}}^{{-6}}})}}{{{{{(1.06)}}^{2}}}}=4\times {{10}^{4}}\text{N}
Illustration
The figure shows a system of two charges. A point charge \displaystyle {{q}_{1}} = 20nC is at (-1,0) while \displaystyle {{q}_{2}} = -10nC is at (+1,0). Find the resultant field intensity at a point P with coordinates (2,2).
Solution
1.
The field vector \displaystyle {{E}_{1}} points away from the charge \displaystyle {{q}_{1}} while the vector \displaystyle {{E}_{2}} points toward the charge \displaystyle {{q}_{2}}.
2.
Let
The distance between P and \displaystyle {{q}_{1}}= \displaystyle {{r}_{1}}
and the distance between P and \displaystyle {{q}_{2}}= \displaystyle {{r}_{2}}
By distance formula
\displaystyle {{r}_{1}} = \sqrt{{{{{(2+1)}}^{2}}\,+{{2}^{2}}}}\,=\sqrt{{13}}=3.6\,m,
\displaystyle {{r}_{2}} = \sqrt{{{{1}^{2}}\,+{{2}^{2}}}}\,=\sqrt{5}\,=\,2.2\,m
The magnitude of the fields are
\displaystyle {{E}_{1}} = \frac{{k{{q}_{1}}}}{{r_{1}^{2}}}\,\,=\,\,\frac{{\left( {9\,\,\times \,{{{10}}^{9}}} \right)\,\,\left( {20\,\,\times \,{{{10}}^{{-9}}}} \right)}}{{{{{(3.6)}}^{2}}}} =\,13.8\,N{{C}^{{-1}}}
\displaystyle {{E}_{2}} = \displaystyle \frac{{k{{q}_{2}}}}{{r_{2}^{2}}}\,\,=\,\,\frac{{\left( {9\,\,\times \,{{{10}}^{9}}} \right)\,\,\left( {10\,\,\times \,{{{10}}^{{-9}}}} \right)}}{{{{{(2.2)}}^{2}}}} \displaystyle =\,18\,\,N{{C}^{{-1}}}
3.
The components of the resultant field are
\displaystyle {{E}_{x}}={{E}_{1}}\cos {{\theta }_{1}}-{{E}_{2}}\cos {{\theta }_{2}}
\displaystyle {{E}_{y}}={{E}_{1}}\sin {{\theta }_{1}}-{{E}_{2}}\sin {{\theta }_{2}}
were \displaystyle \cos {{\theta }_{1}} = \frac{3}{{3.6}} ; \displaystyle \cos {{\theta }_{2}}= \frac{1}{{2.2}}
and \displaystyle \sin {{\theta }_{1}} = \frac{2}{{3.6}} ; \displaystyle \sin {{\theta }_{2}} = \frac{2}{{2.2}}
Thus, Ex = (13.8) \frac{3}{{3.6}}\,-\,18\,\left( {\frac{1}{{2.2}}} \right)\,=\,3.3\,N/C
Ey = (13.8) \frac{2}{{3.6}} – 18 \left( {\frac{2}{{2.2}}} \right) = -8.7 NC–1
Finally, \displaystyle \overrightarrow{E}=3.3\hat{i}-8.7\hat{j}\,N{{C}^{{-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.
{\overrightarrow{{E}_{1}}}, {\overrightarrow{{E}_{2}}}, {\overrightarrow{{E}_{3}}} are the electric field due to +Q, +Q and – Q.
\displaystyle {{\vec{E}}_{1}}=\frac{Q}{{4\pi {{\in }_{0}}{{{\left( {\frac{a}{{\sqrt{3}}}} \right)}}^{2}}}}(\cos 30{}^\circ \hat{i}+\sin 30{}^\circ \hat{j})
\displaystyle {{\vec{E}}_{2}}=\frac{Q}{{4\pi {{\in }_{0}}{{{\left( {\frac{a}{{\sqrt{3}}}} \right)}}^{2}}}}(-\cos 30{}^\circ \hat{i}+\sin 30{}^\circ \hat{j})
\displaystyle {{\vec{E}}_{3}}=\frac{Q}{{4\pi {{\in }_{0}}{{{\left( {\frac{a}{{\sqrt{3}}}} \right)}}^{2}}}}(+\hat{j})
\displaystyle {{\vec{E}}_{{net}}}={{\vec{E}}_{1}}+{{\vec{E}}_{2}}+{{\vec{E}}_{3}}
\displaystyle =\frac{{6Q}}{{4\pi {{\in }_{0}}{{a}^{2}}}}\hat{j}
2nd method
By symmetry we find 2 observation
(1)
All {\overrightarrow{{E}_{1}}}, {\overrightarrow{{E}_{2}}}, {\overrightarrow{{E}_{3}}} have equal magnitude which is given by \frac {kq}{{{r}^{2}}} where r is the distance of point O from any charge and r = \frac{a}{\sqrt{3}}.
(2)
The x components of {\overrightarrow{{E}_{1}}}{ }and{ } {\overrightarrow{{E}_{2}}} being equal and opposite cancels with each other and hence the net electric field is given by the sum of y components of {\overrightarrow{{E}_{1}}}{ },{ } {\overrightarrow{{E}_{2}}} and {\overrightarrow{{E}_{3}}}.
Hence
{{E}_{net}}={{E}_{1}}sin\theta+{{E}_{2}}sin\theta+{{E}_{3}}
where \theta={{30}^{o}}
therefore {{E}_{net}}=\frac{Q}{{4\pi {{\in }_{0}}{{{\left( {\frac{a}{{\sqrt{3}}}} \right)}}^{2}}}}{ }sin{{30}^{o}} + \frac{Q}{{4\pi {{\in }_{0}}{{{\left( {\frac{a}{{\sqrt{3}}}} \right)}}^{2}}}}{ }sin{{30}^{o}} + \frac{Q}{{4\pi {{\in }_{0}}{{{\left( {\frac{a}{{\sqrt{3}}}} \right)}}^{2}}}}
so finally {{E}_{net}}= \frac{{6Q}}{{4\pi {{\in }_{0}}{{a}^{2}}}} along +ve y axis
Objective Assignment
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) \frac{Q}{{4\pi {{\varepsilon }_{o}}{{d}^{2}}}}
(b) \frac{Q}{{36\pi {{\varepsilon }_{o}}{{d}^{2}}}}
(c) \frac{{3Q}}{{4\pi {{\varepsilon }_{o}}{{d}^{2}}}}
(d) \frac{{2Q}}{{9\pi {{\varepsilon }_{o}}{{d}^{2}}}}
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
An oil droplet of density \rho and radius r carries a charge q when placed in a electric field E, its moves upwards with a velocity \nu .If \rho is the density of air, \eta be the viscosity of the air, then which of the following forces is directed upwards ?
(a) 6 \pi \rho \eta r
(b) q E
(c) \frac{4}{3}\pi {{r}^{3}}(\rho -{{\rho }_{0}})g
(d) qE + 6 \pi \rho \eta \nu
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) \displaystyle {{E}_{A}}>{{E}_{B}}
(b) \displaystyle {{E}_{A}}={{E}_{B}}/r
(c) \displaystyle {{E}_{A}}<{{E}_{B}}
(d \displaystyle {{E}_{B}}={{E}_{A}}/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) \frac{1}{{4\pi {{\varepsilon }_{o}}}}\frac{q}{{{{a}^{2}}}}
(b) \frac{q}{{16\pi {{\varepsilon }_{o}}{{a}^{2}}}}
(c) \frac{{\sqrt{2}\,q}}{{4\pi {{\varepsilon }_{o}}{{a}^{2}}}}
(d) \frac{{5q}}{{16\pi {{\varepsilon }_{o}}{{a}^{2}}}}
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 \frac{Q}{{{{r}^{2}}}}
(b) (n-1)k \frac{Q}{{{{r}^{2}}}}
(c) \frac{n}{{n-1}} k \frac{Q}{{{{r}^{2}}}}
(d) \frac{{n-1}}{n} k \frac{Q}{{{{r}^{2}}}}
Ans . (a)
Q.8
Charge Q is given a displacement aî+bĵ in an electric field \displaystyle {{E}_{1}}\hat{i}+{{E}_{2}}\hat{j}. The work done is
(a) \displaystyle Q\left( {{{E}_{1}}a+{{E}_{2}}b} \right)
(b) Q \sqrt{{(2{{E}_{1}}a)+(3{{E}_{2}}b)}}
(c) Q\left( {{{E}_{1}}+{{E}_{2}}} \right) \sqrt{{{{a}^{2}}+{{b}^{2}}}}
(d) Q \displaystyle \sqrt{{(2{{E}_{1}}b+{{E}_{2}}a)}}
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) \displaystyle \frac{1}{{4\pi {{\varepsilon }_{0}}}}\,\,\frac{q}{{3\sqrt{3}{{R}^{2}}}}
(b) \displaystyle \frac{1}{{4\pi {{\varepsilon }_{0}}}}\,\,\frac{{2q}}{{3{{R}^{2}}}}
(c) \displaystyle \frac{1}{{4\pi {{\varepsilon }_{0}}}}\,\,\frac{{2q}}{{3\sqrt{3}{{R}^{2}}}}
(d) \displaystyle \frac{1}{{4\pi {{\varepsilon }_{0}}}}\,\,\frac{{3q}}{{2\sqrt{2}{{R}^{2}}}}
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 {{m}_{e}}, initially at rest, moves through a certain distance under a uniform electric field in time {{t}_{e}}. A proton of mass {{m}_{p}} also initially at rest, takes time {{t}_{p}} to move through the same distance in the same electric field. Neglecting the effect of gravity, the ratio \frac {{t}_{e}}{{t}_{p}} is nearly equal to
(a) 1
(b) \displaystyle \sqrt{{{{m}_{p}}/{{m}_{e}}}}
(c) \displaystyle \sqrt{{{{m}_{e}}/{{m}_{p}}}}
(d) 1836
Ans. (c)
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) \frac{L}{4}
(c) 8 L
(d) 4 L
Ans (a)
Subjective Assignment
Q.1
Find the electric field required to balance the weight of an (1) electron (2) alpha particle.
.Ans: 56.8 \times {{10}^{-11}}, 2 \times {{10}^{-7}}{ }N/C
Q.2
An e is moving with a speed of 104 m/sec enter a electric field of 1200N/c the electric field is applied in opposite direction of the motion of e Find the time in which it attain the velocity of light?
Ans: 1.4 \displaystyle \mu sec
Q.3
A simple pendulum consists of a bob of mass 10gm and is suspended vertically from a fixed point. If suddenly an electric field of 100N/C is applied horizontally around the pendulum the string of the pendulum starts making 30o with the initial position. Find the charge over the bob.
.Ans: 0.57 \times {{10}^{-3}} C
Q.4
A pendulum bob of mass 80 milligram and carrying a charge of 2×10-8 C is at rest in a horizontal uniform electric field of 2×104 Vm-1. Find the tension in the thread of the pendulum and the angle it makes with the vertical.
Ans: 8.801×10-4 N
Q.5
An electron moving with a speed of 5.0 x108 cm/sec is shot parallel to an electric field of strength 1.0 x 103 N/C arranged so as to retard its motion.
(a)
How far will the electron travel in the field before coming (momentarily) to rest
Ans: 0.07 m
(b)
How much time will elapse? if the electric field ends abruptly after 0.8 cm, what fraction of its initial energy will the electron lose in traversing it?
Ans: 0.11
Q.6
An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55×104 Vm-1 in Millikan oil drop experiment. The density of the oil is 1.26 gm cm-3. Estimate the radius of the drop.
Ans: 9.81×10-7m
Q.7
When a test charge q0 = 2 nC is placed at the origin, it experiences a force of 8.0×10-4 N in the positive y-direction. What would be the force on a charge -4 nC placed at the origin?
Ans: -16×10-4 N
Q.8
A particle of mass m and charge q is thrown at a speed u against a uniform electric field E. How much distance will it travel before coming to momentary rest?
Ans: mu2 / 2qE
Q.9
The path followed by an electric charge in an electric field when moved perpendicular to it is a parabola. Prove it.
Q.10
Two parallel plates have an electric field of 200 N/c directed upwards.An electron is projected at a speed of 6×106 m/sec at an angle of 30o .from the lower plate from the origen (0,0).Distance between the plates is 10 cm .
(a) Will the electron strike the upper plate.
Ans: Yes
(b) If it strikes the plate what is the approx. coordinates of that point?
Ans: (23,10)cm
Q.11
Two oppositely charged horizontal parallel plates are kept 30 cm apart with an electric field of 10N/C between them .A charge particle of mass 1gm and charge 2mC is projected horizontally with a speed of 30m/sec between them .If the length of the plates are 30cm each then find the deflection of the electron by the time it leave the electric field.
Ans: 0.1 cm
Q.12
An electron moves in a vertical plane with a speed of 1.41×107 ms-1 and enters a region where there is a uniform electric field of 3500 NC-1, directed downwards. Find the co-ordinates of the electron 5×10-8 s after if passes through the point of entry along a course directed at an angle of 30° below the horizontal.
Ans: 0.61 m and 0.42 m
Q.13
Two point charges qA = 3\mu C and qB= – 3\mu C are located 20- cm apart in vacuum
(a)
What is the electric field at the mid point O of the line AB joining the two charges?
Ans: 5.4 \times {{10}^{6}} N/C
(b)
If a negative test charge of magnitude 1.5×10-9 C is placed at a point 20cm from both the charges , what is the force experienced by the test charge?
Ans: 1.75 \times {{10}^{-3}} N
Q.14
Three charges of 3\mu C each are kept at corners of a square of side 1m. Find the net electric field at the center of square.
Ans: 54\times {{10}^{3}}{ }N/C
Q.15
Find the electric field vector at (0,0,5)m due to two charges of 1mC at (0,4,0) and -2mC at (3,0,0).
Ans: (0.27\hat{i}-0.135\hat{j}-0.279\hat{k})\times {{10}^{6}}N/C
Q.16
Four particles, each having a charge q is placed on the four corners A, B, C, D of a regular pentagon ABCDE. The distance of each corner form the centre is a. Find the electric field at the centre of the pentagon.
Ans: \frac{q}{{4\pi {{\varepsilon}_{o}}{{a}^{2}}}} along OE
Q.17
An infinite number of charges, each equal to q are placed along the x-axis , at x= 1, x = 2, x = 4, x=8, x= 16, and so no. Find the electric field, at point, x = 0, due to this set of charges.
Ans: E=\frac{q}{{3\pi {{\varepsilon}_{o}}}}
Q.18
Two point charges of unknown magnitude and sign are placed a distance d apart.
(a) If it is possible to have E= 0 at any point not between the charges but on the line joining them, what are the necessary conditions and where is the point located?
Ans: When the charges are of opposite sign the neutral point exists outside the system on the line joining them near the charge which is smaller in magnitude.
(b) Is it possible, for any arrangement of two point charge, to find two points (neither at infinity) at which E = 0; if so, under what conditions?
Ans: No
Q.19
Two point charges of 2mC and 8mC are placed 12 cm apart. Find the position of the point where the electric field intensity will be zero.
Ans: 4 cm from 2mC charge
Q.20
Eight identical point chares of q coulomb each are placed at the corners of a cube of each side 0.1 m. Calculate electric field at the centre of the cube. Calculate the field at the centre when one of the corner charges is removed.
Ans: zero , 1.2\times {{10}^{12}}{ }N/C q NC-1 towards the corner without charge
Q.21
A charge of 4×10-9 C is distributed uniformly over the circumference of a conducting ring of radius 0.3 m. Calculate the field intensity at a point on the axis of the ring at 0.4 m from its centre, and also at the centre.
Ans: 115.2 N/C ; Zero
Q.22
A thin rod of length L has a uniform linear charge density \lambda . Find the electric field at a distance r on the right bisector of this rod.
Ans: \displaystyle \frac{{\lambda L}}{{2\pi {{\varepsilon }_{o}}r\sqrt{{{{L}^{2}}+4{{r}^{2}}}}}}
Q.23
Two oppositely charged ri ngs of linear charge density 10C/m & radius of 10cm each are kept coaxial at 20 cm from each other. If an electron is kept midway between them find the force acting on it.
Ans: 4.8×10-7 N
Q.24
A line of infinite length and of linear charge density 100C/m is kept along y axis. A proton is released from rest from a point on this line. Find the distance it will travel before reaching a velocity of 1000m/sec.
Ans: Approx 1m
Q.25
(a)
Is the electric field due to a charge configuration with total charge zero, necessarily zero ? Justify.
Ans: No for example the net charge on DIPOLE is zero but the electric field is not zero.
(b)
Figure shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?
Ans : 3rd charge has maximum charge to mass ratio as its deflection is maximum
