In this page we have Multiple Choice questions on Electric Charge for Jee Main and Advanced . Hope you like them and do not forget to like , social share
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Question 1
A charge of magnitude q is divided into two parts such that force between resulting two charges is maximum when separated through some distance r. The division of charges would be
(a) 3q/8 , 5q/8
(b) 2q/4 , 2q/4
(c) q/2 , q/2
(d) 3q/6 , 3q/6 Solution
Suppose one part of charge is $q_1$ then second part is $(q-q_1)$ Therefore
$F= \frac {1}{4 \pi \epsilon _0} \frac {q_1(q-q_1)}{r^2}$
For F to be maximum $\frac {dF}{dr} =0$ , this gives $q_1=\frac {q}{2}$
Hence (c) is correct
Question 2
Consider a system of three charges q/3, q/3 and –2q/3 placed at points A, B and C respectively as shown in the figure. Take O to be the centre of the circle of radius R and angle CAB = 60º
(a) The electric field at point O is q/8πε_{0}R^{2}
(b) The magnitude of the force between the charges at C and B is q^{2}/54πε_{0}R^{2}
(c) The potential energy of the system is zero
(d) The potential at point O is q/12πε_{0}R Solution
The electric field at O is due to the negative charge at C only since the equal positive charges situated at A and B will produce equal and opposite fields at O (and they will mutually cancel). The field at O is therefore negative and the option (a) is obviously wrong. Option (c) also is obviously wrong.
The magnitude of the force (F) between the charges at C and B is given by
Thus,
$F= \frac {1}{4 \pi \epsilon _0} \frac {(2q/3)(q/3)}{(2R \sin 60)^2}$
or
$F= \frac {q^2}{54 \pi \epsilon _0 R^2}$
Hence (b) is correct
Question 3
Four charges q, 2q, 3q, 4q are placed at corners A, B, C and D of a square as shown below in the figure. The field at centre O of square has the direction along
We know that electric field at a distance r due to a point charge Q is given by
$E= \frac {Q}{4 \pi \epsilon _0 r^2}$
We also know that field of a positive charge is directed away from the charge. From the figure given above we can figure out that OD=OC=OA=OB. Electric field at O due to charge 2q at B is
$E_{BO} =\frac {2q}{4 \pi \epsilon _0 r^2}$ along OD
Electric field at O due to charge 4q at D is
$E_{DO} =\frac {4q}{4 \pi \epsilon _0 r^2}$ along OB
Therefore, net field along OB is $E=E_2-E_1$ which is equal to
$E=\frac {2q}{4 \pi \epsilon _0 r^2}$
Similarly, net electric field at point O due to charges q and 3q at A and C is
$E^{'}= \frac {2q}{4 \pi \epsilon _0 r^2}$ directed along OA
Thus E=E' , but they are mutually perpendicular to each other, therefore their resultant would be along PQ which is parallel to CB.
Hence (b) is correct
Question 4
A point charge q is placed at geometrical centre of one of the face of a cube. The total flux through the cubical surface due to charge is
(a)$\frac {q}{\epsilon _0}$
(b) $\frac {q}{2\epsilon _0}$
(c) $\frac {2q}{\epsilon _0}$
(d) 0 Solution
Consider the charge q is placed at the centre of face ADHE of the cube shown below in the figure
If we construct an identical cube to the left of this cube such that the face ADHE is common to both cubes, we then have a closed Gaussian surface if we remove the common face ADHE. Now applying symmetry considerations, we find that flux through each cube is half of the total value $\frac {q}{\epsilon _0}$
Hence (b) is correct
Question 5
Two large metal sheets having surface charge density +σ and –σ are kept parallel to each other at a small separation distance d. The electric field at any point in the region between the plates is
(a) σ/ε_{0}
(b) σ/2ε_{0}
(c) 2σ/ε_{0}
(d) σ/4ε_{0} Solution
Consider the figure given below.
Left plate produces field $ \frac {\sigma}{2\epsilon _0}$ which points away from it to left in region (i) and to right in region (ii) and (iii). The negatively charged right plate produces field $ \frac {\sigma}{2\epsilon _0}$ which points towards it , to the right in region (i) and (ii) to left in region (iii). The two fields cancels in region (i) and (iii) and the field is $ \frac {\sigma}{\epsilon _0}$ and points to the right in region between the plates.
Hence (a) is correct
Question 6
A rod lies along the x-axis with one end at the origin and other at x->∞ it caries a uniform charge λ C/m. Find the electric field at the point x=-a on the x-axis
(a) $\frac {-\lambda}{4 \pi \epsilon _0 a} \boldsymbol{i}$
(b) $\frac {-\lambda}{4 \pi \epsilon _0 a^2} \boldsymbol{i}$
(c) $\frac {\lambda}{4 \pi \epsilon _0 a} \boldsymbol{i}$
(d) $\frac {\lambda}{4 \pi \epsilon _0 a^2} \boldsymbol{i}$ Solution
Because of the location of the field point $\boldsymbol{E}=E(-\boldsymbol{i})$
Now consider a small length dx on the line charge
Electric field due to this element at $A= \frac {\lambda dx}{4 \pi \epsilon _0 ( a+x)^2}$
Electric field due whole length of charge
$E = \frac {\lambda}{4 \pi \epsilon _0} \int_{0}^{\infty } \frac {dx}{(a+x)^2}$
Calculating
$E= \frac {\lambda}{4 \pi \epsilon _0 a}$
or
$\boldsymbol{E}=\frac {-\lambda}{4 \pi \epsilon _0 a} \boldsymbol{i}$
Question 7
Twelve charges of charge q are situated at the corners of the 12 sided polygon of side a. What is the net force on the charge Q at the centre
(a) Zero
(b) 3qQ/πε_{0}a^{2}
(c) qQ/πε_{0}a^{2}
(d) None of the above Solution
Due to symmetry all force due to charges cancel each other at centre
Question 8
Two positive point charge are placed at the distance a apart have sum Q. What values of the charges , coulomb force between them is maximum
(a) q_{1}=q_{1}=Q/2
(b) q_{1}=3Q/4 ,q_{2}=Q/4
(c) q_{1}=5Q/6 ,q_{2}=Q/6
(d) Non of the above Solution
let q be the one charge
then Q-q be the other charge
Force between them
$F=\frac {q(Q-q)}{4 \pi \epsilon _0 a^2}$
$\frac {dF}{dq}=\frac {(Q-2q)}{K}$ Where $K =4 \pi \epsilon _0 a^2$
for dF/dq=0
$q=\frac {Q}{2}$
Also
$\frac {d^2F}{dq^2}=-2$
So, force of attraction got maximum at q=Q/2
Hence (a) is correct
Question 9
A metallic shell having inner radius R_{1} and outer radii R_{2} has a point charge Q kept inside the cavity. Electric field in the region R_{1} < r < R_{2} where r is the distance from the centre is given by
(a) depends on the value of r
(b) Zero
(c) Constant and nonzero everywhere
(d) None of the above Solution
Electric field inside the conductor is zero as electron move in such a way in the conductor so as to nullify any electric field produced by the charge
So answer is b
Question 10
Consider two statements (A) The force with which two charges interact is not changed by the presence of the other charges (B) Electric force experienced by the charge particle due to number of fixed point charges is vector resultant of the forces experience due to individual charges
(a) A and B both are correct
(b) A is correct only
(c) B is correct only
(d) A and B both are wrong Solution
(A)The force with which two charges interact is not changed by the presence of the other charges: it is correct we know this from coulomb law
(B)Electric force experienced by the charge particle due to number of fixed point charges is vector resultant of the forces experience due to individual charges: this is also correct from law of superimposition of coulomb forces
Question 11
A metallic solid sphere of radius R is given the charge Q. Which of the following statement is true then
(a) Electric field at points 0< r < R is zero
(b) Charge Q is on the outer surface of the sphere
(c) Electric field at r>R is given by Q/4πε_{0}r^{2}
(d) Electric field is perpendicular to the surface of the sphere Solution
We know that electric field inside the conductor is zero as electron move in such a way in the conductor so as to nullify any electric field produced by the charge Now since no electric field is there inside the sphere, flux through a spherical surface inside the sphere must be zero. So, from Gaussian law net charge inside the sphere is zero. So all charge given to the sphere resides on the surface of the sphere.
From symmetry it can be said that Electric Field must be same on a spherical surface concentric to solid sphere outside the sphere.
So applying Gaussian law to that surface
$E \times ? 4 \pi r^2=\frac {q}{\epsilon _0}$
or $E=\frac {q}{4 \pi \epsilon _0 r^2}$
Also electric field is perpendicular to the surface as otherwise there would an movement of charge on the surface
Hence (a),(b),(c),(d) are correct
Question 12.
A simple pendulum consists of a small sphere of mass and positive charge q is suspended by the string of length L. The pendulum is placed in the electric field of strength E directed vertically downwards. Which of the following is true
(a)Time period of oscillation= $T= 2 \pi \sqrt { \frac {L}{g+qE/m}}$
(b)Time period of oscillation=$T= 2 \pi \sqrt { \frac {L}{g+qE/m}}$
(c) Tension in the string when the pendulum is at rest =$mq+qE$
(d)Tension in the string when the pendulum is at rest =$mq-qE$ Solution
Two forces are acting on the charge
Weight mg vertically downwards
Electrical force due to electric field present qE vertically downwards
So net downwards force=mg+qE
So net acceleration due to force=g+qE/M
We know that Time period for simple pendulum
$T=2 \pi \sqrt {\frac {L}{g}}$
where g is acceleration downwards
So in this case
$T=2 \pi \sqrt {\frac {L}{g+qE/m}}$
Also when the pendulum is at rest
Tension is the string provide the upward pull to prevent the charge from falling
So tension=$mg+qE$
Hence (a),(c) are correct
Question 13
A rod lies on the x-axis with end and at x=-L and other end at x=L with uniform charge λ C/m. Which of the following is true. here $k=\frac {1}{4\pi \epsilon _0}
(a).Electric field at any point (0,y) on the y-axis is given by E=(2kλL/y√(y^{2}+L^{2}))j
(b) For point on the Y-axis greater than y>>L E=(2kλL/y^{2})j
(c) Electric field if L->∞ E=2kλ/y
(d) None of the above Solution
It is advantageous to chop the line up into symmetrically placed pairs , for then the horizontal components of the two field cancel and the net field of the pair is
$dE=2( \frac {1}{4 \pi \epsilon _0}) ( \frac {\lambda dx}{r^2}) \cos \theta y$
here $ \cos \theta =\frac {y}{\sqrt {y^2+x^2}}$ and x runs from 0 to L
$E=\frac {\lambda}{4 \pi \epsilon _0} \int \frac {2ydx}{\sqrt {y^2+x^2}}$
$E= \frac {2\lambda L}{4 \pi \epsilon _0} \frac {1}{y \sqrt {y^2+L^2}}$
And it direction is along y axis. For points far from the line (y>>>L)
then
$E=(\frac {1}{4 \pi \epsilon _0})(\frac {2 \lambda L}{y^2})$
For L->∞,it is equivalent to infinite straight wire
$E=\frac {2k \lambda}{y}$
Hence (a),(b),(c) are correct
Question 14
A particle of mass m and charge q is thrown horizontally with a velocity v from top of the building of height H. An electric field exists in the plane and it is horizontally away from the building
which of the following is true
(a) Range of the particle is greater than v√(2H/g)
(b) Time of flight is √(2H/g)
(c) Path is parabolic
(d) None of the above Solution
Initial vertical component of velocity=0
Vertical component of the velocity=v
Vertical component of acceleration downwards=g
Horizontal components of acceleration away from the build=qE/m
Vertical motion
$H= \frac {1}{2}gt^2$
or $t= \sqrt {\frac {2H}{g}}$
So time of flight is $\sqrt {\frac {2H}{g}}$
Horizontal motion
$s=vt+\frac {1}{2}( \frac {qE}{m})t^2$
$ s=v \sqrt {\frac {2H}{g}} +{1}{2}( \frac {qE}{m})(\frac {2H}{g})$
So range is greater then $v \sqrt {\frac {2H}{g}}$
For horizontal motion
$y=\frac {1}{2}gt^2$
$x=vt+\frac {1}{2} (\frac {qE}{m})t^2$
or
$x=v \sqrt {\frac {2y}{g}}+\frac {1}{2}(\frac {qE}{m})(\frac {2y}{g})$
or $x=v \sqrt {\frac {2y}{g}} + \frac {qEy}{mg}$
which is not a parabola
Hence (a),(b) are correct
Question 15
At a point on the axis of an electric dipole
(a) Electric field is zero
(b) Electric potential is zero
(c) Neither electric field nor electric potential is zero
(d) Electric field is directed perpendicular to axis Solution
(c) is correct
Question 16
A point charge (Q) is located at the centre of a cube of edge length a, find the final electric flux over one face of the cube
a. Q/ε_{0}
b. Q/6ε_{0}
c. 6Q/ε_{0}
d. none of the above Solution
flux will be same for all faces
So applying Gauss law for this
$\int \mathbf{E}.d\mathbf{S}= \frac {q}{\epsilon _0 }$
$6 \times \text{flux of one face} = \frac {Q}{\epsilon _0 }$
$\text {flux of one face} = \frac {Q}{6\epsilon _0 }$
hence (b) is correct
Question 17
Three point charges q + Q, q, q - Q are enclosed by the surface S. What the net flux crosses S
a. 3q
b. 2q
c. 3q - Q
d. can not be determine based on the data given in question Solution
From Gauss law
$\int \mathbf{E}.d\mathbf{S}= \frac {q_{in}}{\epsilon _0 }$
So Flux=q+Q+q+q-Q=3q
Hence (a) is correct
Question 18
Find the electric field inside the sphere which carries a charge density proportional to the distance from the origin
ρ = kr
a. ρ/ε_{0}
b. ρr/ε_{0}
c. ρr^{2}/ε_{0}
d. none of the above Solution
Draw a Gaussian surface at distance r from surface, by symmetry E is uniform at every point
$\int \mathbf{E}.d\mathbf{S}= \frac {q_{in}}{\epsilon _0 }$
Now
$q_{in} = \int_{0}^{r} dq= \int_{0}^{r} \rho \times 4 \pi x^2 dx=\int_{0}^{r} k x \times 4 \pi x^2dx =4 \pi k \int_{0}^{r} x^3 dx= \pi kr^4$
Therefore
$E \times 4 \pi r^2 = \frac {\pi kr^4}{\epsilon _0 }$
$E= \frac {kr^2}{\epsilon _0 }$
$E= \frac {\rho r}{\epsilon _0 }$
hence (b) is correct
Question 19
A point charge Q(C) is placed at the origin. Find the electric flux of which an area 4π m^{2} on a concentric spherical shell of radius R
a. Q/R^{2}ε_{0}
b. Q/ε_{0}
c. Q/4R^{2}ε_{0}
d. none of the above Solution
Total flux = $\frac {Q}{\epsilon _0 }$
Flux per unit Area= $\frac {Q}{4 \pi R^2 \epsilon _0 }$
Flux on $4 \pi$ = $(\frac {Q}{4 \pi R^2 \epsilon _0 }) \times 4 \pi = \frac {Q}{R^2 \epsilon _0 }$
hence (a) is correct
Question 20
As per Gauss law
Which of the following is true about this
a. This is valid for symmetrical surface only
b. E is the electric field to the charge inside the surface
c. Electric flux on the closed surface due to outside charge is always zero
d. none of the above Solution
(c) is correct
Question 21
A uniform line charge with linear density λ lies along the y-axis. What flux crosses a spherical surface cantered at the origin with r = R
a. 2Rλ/ε_{0}
b. Rλ/ε_{0}
c. λ/ε_{0}
d. none of the above Solution
Total charge inside the spherical surface =2Rλ
$\int E.dS$ = q_{in}/ε_{0}
So flux = 2Rλ/ε_{0}
Hence answer is (a)
Linked comprehension type (A)
A thin metallic wire ring of radius R and area of cross-section A carries an electric charge of q Coulombs. The centre of the ring contains a charge Q>>q with sign of Q same as that of q. Question 22
The force with which ring stretches is
Question 23
Change in radius of the ring would be
where Y is the young’s modulus of the ring Solution 22-23
Consider a small element of the ring as shown below in the figure
Let length of this element of ring is $R \alpha$. Now force exerted by charge Q on this element is
$\Delta F = \frac {Q \Delta q}{4 \pi \epsilon _0 R^2}$
Where
$\Delta q = \frac {q \alpha}{2 \pi}$
This force $\Delta F$ is balanced by the tension of the ring which pulls tangentially at each end of the segment. The horizontal components of T,T cancels each other but vertical components add up to form a restoring force which balances $\Delta F$. So
$\Delta F = 2T \sin \frac {\alpha}{2} \approx T \alpha $ since $\alpha$ is very small
$ T = \frac {\Delta F}{\alpha} = \frac {1}{ 4 \pi \epsilon _0 } \frac {Qq \alpha}{2 \pi R^2 \alpha} = \frac {1}{ 4 \pi \epsilon _0 } \frac {qQ}{2 \pi R^2}$
which is the force with which the ring is stretched.
The longitudinal strain in the circumference is
$\text{strain}= \frac {\text{stress}}{\text {youngs modulus}} = \frac {T}{YA} $
Change in circumference = strain × circumference =$\frac {T}{YA} \time 2\pi R$
Now
Change in Radius is given
$\Delta R = \frac {\text{Change in circumference}}{2 \pi} =\frac {TR}{AY} = \frac {1}{ 4 \pi \epsilon _0 } \frac {qQ}{2 \pi RAY}$
Hence Answer for question 22 is (a) and question 23 is (d)
Linked comprehension type (B)
A slab of uniform thickness 2d and uniform charge density ρ is lying between –d to d along x-axis and extends infinitely along y and z directions as shown below in the figure.
Question 24
Magnitude of electric field at any point inside the slab as a function of x is
Question 25
Electric field for regions outside the slab is
Electric field on yz plane is zero by symmetry. Draw a Gaussian pill box extending from center and below the surface of the slab. Applying gauss's law for the region inside the slab we have
$\int \mathbf{E}.d\mathbf{S}= \frac {q_{enc}}{\epsilon _0 }$
$EA= \frac {q_{enc}}{\epsilon _0 }$
If $\rho$ is the charge per unit volume in the slab then charge inside the pillbox is
$Q_{enc} = \rho A x$
or,
$EA= \frac {\rho A x}{\epsilon _0 }$
or, $E=|E|=\frac {\rho x}{\epsilon _0 }$
This field points away from the surface of the slab
For region outside the slab,
$Q_{enc} = \rho A d$
On applying Gauss's law we find
$\mathbf{E}= \frac {\rho d}{\epsilon _0}\mathbf{i} $
Note: We have consider positive direction for solving this problem
Hence Answer for question 24 is (a) and question 25 is (c)
Linked comprehension type (C) Question 26
A long cylinder contains charge distributed uniformly having volume charge density ρ. (A) Electric field inside this cylinder is
First draw a Gaussian cylinder of length l and radius x inside the given long cylinder as shown below in the figure.
Question 27
What would be the field inside the cylinder if volume charge density of the cylinder varies according to the following relation given below?
ρ=kx^{2}
where k is a constant.
Assertion reason type questions
The following questions consist of two statements, Statement I and Statement II. While answering these questions choose any of the following four responses
A. If both Statement I and Statement II are true and the reason is correct explanation of Statement I.
B. If both Statement I and Statement II are true but reason is not a correct explanation of Statement I.
C. If Statement I is true and Statement II is false.
D. If both Statement I and Statement II are false. Question 28 Statement I: There is no electric field in the conductor Statement II: There are plenty of free electrons in the conductors which moves in such a way, it cancel every electric field in the conductor Solution
Statement I is true
Statement II is also true
Statement II is correct explanation for statement I
Question 29 Statement I: Coulombs Force is an action reaction pair Statement II: Coulombs force is non conservative in nature Solution
Statement I is true
Statement II is false as it is a conservative force
Question 30 Statement I: Two electric lines of force can never intersect Statement II: Electric lines of force originates from negative and terminate on positive charge Solution
Statement I is true
Statement II is false as it is a conservative force
Question 31 Statement I: Excess charge given to the conductor always resides on the outer surface of the conductor Statement II: electric field is zero inside the bulk material of the conductor Solution
Statement I is true
Statement II is also true
Statement II is correct explanation for statement I
Matrix Match question Question 32
Consider the figure given below
A charge Q is placed inside the Spherical conductor shell Column I
(A) Charge Q is at the center of the shell
(B) Charge Q is displaced from center and place at distance r from the center
(C) An excess charge q is given to the shell
(D) A charge q is placed near the conductor shell Column II
(X) Charge distribution is uniform at the inner surface of shell
(Y) Charge distribution is non uniform at the inner surface of shell
(Z) Charge distribution is uniform at the outer surface of shell
(K) Charge distribution is non uniform at the outer surface of shell Solution
When the charge is placed inside the shell, an induced charge -Q will be there on inner shell and +Q will be present at the outer shell.
When the charge is present at the center, Charge would be uniformly distributed on the inner surface and on the outer surface.
When the charge is present at some distance from center, Charge would be none uniformly distributed on the inner surface. But it will be uniformly distributed over the outer surface
When an excess charge is given to shell, it will reside on the surface of the shell and distribute uniformly over the shell. Inner shell will not have any effect of it.
When a charge q is placed near the conductor shell, charge distribution on the outer shell will change because of the electric field of the outside charge. There will be no effect to the inner surface.
So answer is
A -> X ,Z
B -> Y,Z
C -> X,Z
D -> X,K
Question 33
Consider the figure given below in which an infinite conductor has uniform surface charge distribution σ_{0} and adjacent to it is an infinite parallel layer of charge with volume charge density ρ varying according to following relation
ρ=ρ_{0}e^{-αz} , where z>0
Along x and y directions electric field is perpendicular to conductor and hence does not contribute. Electric field is not uniform along z axis (as can be seen from the figure) because of the varying volume charge density. Applying gauss's law to surface S of the figure we see that charge enclosed depends on the z dimensions of the box. From Gauss's Law
Question 34
Two uniform infinite sheets of charge density +σ and –σ intersect at right angles. Electric field due to this configuration at any point in space is
Consider an infinite sheet having charge density +s. Magnitude of electric field due to this sheet of charge is
$E=\frac {\sigma }{2\epsilon _0 }$
The direction of electric field is perpendicular to the surface of the sheet. For two orthogonal sheets of charge density $\pm \sigma$ , superposition principle yields
$E=\frac {\sqrt {2} \sigma }{2\epsilon _0 }$
Question 35
A charge Q sits at the back corner of a cube. What is the flux of electric field through the shaded surface?
Think of this cube as one of 8 surrounding the charge. Each of the 24 squares which make us the surface of this large cube gets the same flux as every other one
$\int _{one \; surface} \mathbf{E}.d\mathbf{A}=\frac {1}{24} \int _{full \; cube} \mathbf{E}.d\mathbf{A} $
or
$\phi = \frac {Q}{24\epsilon _0 }$
Question 36 Match the column
A point charge is placed at the center of a closed spherical surface.
Φ : Electrical Flux through the closed spherical surface
E : Electric Field through the closed spherical surface
Then Mark out the correct statement Column I
(A) If the point charge is displaced from the center but kept within the closed surface
(B) If the point charge is displaced such that it crosses the boundary of closed surface
(C) If any another charge is kept outside the surface
(D) If any another charge is placed inside the surface Column II
(P) Φ will change
(Q) E will change
(R) Φ will not change
(S) E will not change Solution
As per Gauss Law
$\int \mathbf{E}.d\mathbf{S}= \frac {q_{enc}}{\epsilon _0 }$
Where E is the net electric field at the Gaussian surface .It will have effect of both inside and outside charges
If the point charge is displaced from the center but kept within the closed surface, As per gauss law, Flux will not change. But electric field will change at the surface
If the point charge is displaced such that it crosses the boundary of closed surface, As per gauss law, Flux will change and electric field will also change at the surface
If any another charge is kept outside the surface, as per gauss law, Flux will not change. But electric field will change at the surface as another charge present will effect the electric field intensity
If any another charge is placed inside the surface. As per gauss law, Flux will change. Electric field will also change as another charge present will effect the electric field intensity
A -> Q, R
B -> P, Q
C -> Q, R
D -> P,Q
Question 37
A point charge +Q is at the origin of a coordinates system . It is surrounded by a concentric uniform distribution of charge on a spherical shell at r=R for which the total charge is -2Q
Which of the following is correct?
Question 38
Two infinite uniform line charges of linear charge density 2 C/m lies in a XY plan at x =+a m and x=-a m.. A positive charge particle of charge q is placed at the origin and it is displaced by a distance k from origin towards +x axis.
Which of the following is correct
(a) The charge particle will oscillate but not harmonically for all values of 0 < k <a
(b) The charge particle will oscillate harmonically for all values of 0 < k <a
(c) The charge particle will oscillate harmonically for k<<<a
(d) The charge particle will not oscillate Solution
Question 39
Two infinite uniform sheet of charge each with surface charge density σ are located at x=+a and x=-a.
Which of the following is correct?
Electric field near the uniform charge sheet is given by
$E=\frac {\sigma}{2\epsilon _0 }$
And direction is away from positive sheet
Now electric field for each region is shown the in figure
So, it is clear
$\boldsymbol{E} = \frac {- \sigma}{\epsilon _0 } \boldsymbol{i}$ for x < -a
$\boldsymbol{E} = 0$ for -a < x < a
$\boldsymbol{E} = \frac { \sigma}{\epsilon _0 } \boldsymbol{i}$ for x > a
Option (a) ,(b),(c) are correct
Question 40
A conductor of non uniform curvature is given Q charge. Which of the following is correct
(a) The charge is distributed uniformly over its volume
(b) The charge is distributed uniformly over its outer surface
(c) The charge has the greatest concentration on the part of greatest curvature
(d) The charge has the greatest concentration on the part of least curvature Solution
Answer is (c)
Question 41
At a point on the axis of an electrical dipole
(a) Electric field is zero
(b) Electric potential is zero
(c) Neither Electric field and electric potential is zero
(d) Electric field is directed Perpendicular to the axis Solution
Answer is (c)
Question 42
A positive charge Q is uniformly distributed along a circular ring of radius R. A small test charge q is placed at the centre of the ring
Then
(a) If q > 0 and is displaced away from the centre in the plane of the ring, it will be pushed back towards the centre.
(b) If q < 0 and is displaced away from the centre in the plane of the ring, it will never return to the centre and will continue moving till it hits the ring.
(c) If q < 0, it will perform SHM for small displacement along the axis.
(d) q at the centre of the ring is in an unstable equilibrium within the plane of the ring for q > 0
Solution
Answer is (a),(b),(c) and (d)
Question 43
A Gaussian surface of radius R with Q at the centre is drawn in the given region of charges.
Which of the following is true
(a) total flux through the surface of the sphere is $ \frac {-Q}{\epsilon _0 }$
(b)field on the surface of sphere due to -2Q is same everywhere
(c) field on the surface of the sphere is $ \frac {-Q}{4 \pi \epsilon _0 R^2}$
(d) flux through the surface of sphere due to 5Q is zero Solution
(a) and (d) from Gauss laws
Question 44
Consider the charge configuration and spherical Gaussian surface as shown in the figure. When calculating the flux of the electric field over the spherical surface the electric field will be due to
(a) $q_2$
(b) Only the positive charges
(c) All the charges
(d) $-q_1$ and $+q_1$ Solution
Answer is (c)
The electric field is due to all charges present whether inside or outside the given surface.