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
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
Consider the figure given below
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
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
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
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
$\boldsymbol{E}=\frac {-\lambda}{4 \pi \epsilon _0 a} \boldsymbol{i}$
Due to symmetry all force due to charges cancel each other at centre
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
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
(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
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
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
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
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
(c) is correct
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
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
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
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
(c) is correct
Total charge inside the spherical surface =2Rλ
$\int E.dS$ = qin/ε0
So flux = 2Rλ/ε0
Hence answer is (a)
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)
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)
First draw a Gaussian cylinder of length l and radius x inside the given long cylinder as shown below in the figure.
Statement I is true
Statement II is also true
Statement II is correct explanation for statement I
Statement I is true
Statement II is false as it is a conservative force
Statement I is true
Statement II is false as it is a conservative force
Statement I is true
Statement II is also true
Statement II is correct explanation for statement I
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
Consider the figure given below
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
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
$E=\frac {\sqrt {2} \sigma }{2\epsilon _0 }$
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
$\phi = \frac {Q}{24\epsilon _0 }$
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
Applying gauss law, a and b option are 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
Answer is (c)
Answer is (c)
Answer is (a),(b),(c) and (d)
(a) and (d) from Gauss laws
Answer is (c)
The electric field is due to all charges present whether inside or outside the given surface.
