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Class XII
Unit-1
Electrostatics
2010
Q.
AnsStable position of the dipole: parallel to electric field
Unstable position: perpendicular to the electric field
Q.
Ans.Potential.
Scaler
Q.
Ans. Electric flux through the surface S will be as per Gauss law:
2011
Q.
Ans. Potential at a point : V = kq/r
For any Q,
Where rA<rB
So 1/rA>1/rB
So
>0
If Q at O is positive, VA-VB will be positive
If Q at O is negative, VA-VB will be negative.
Two uniformly large parallel thin plates having charge densities + and
- are kept in the X-Z plane at a distance 'd' apart. Sketch an
equipotential surface due to electric field between the plates. If a
particle of mass m and charge '-q' remains stationary between the
plates, what is the magnitude and direction of the field?
Or
Two small identical electrical diploes AB and CD, each of dipole
moment 'p' are kept an angle of 120o as shown in the figure. What is
the resultant dipole moment of this combination? If this system is
subjected to electric field
directed along +X direction, what will be
the magnitude and direction of the torque acting on this?
[Marks:2]
Ans Here the red dotted lines represent the parallel equipotential surfaces along X-Z
plane.
If a charge q has to be held stationary between the two plates, it will have to be
balanced by two forces.
Gravitational force: mg, downwards
Here the red dotted lines represent the parallel equipotential surfaces along X-Z
plane.
If a charge q has to be held stationary between the two plates, it will have to be
balanced by two forces.
Gravitational force: mg, downwards
Electrostatic force= 2qE, acting upwards.
This implies, that in X-Z plane, the upper plate is + charged plate and lower plate is charged plate.
So, E field lines have to be directed along -y axis.
Or
Resultant dipole moment,
Direction of resultant dipole moment :
That is, 30 degrees with +x axis.
Given applied E is along +x axis,
So torque on resultant dipole will be
Direction will be along -z axis.
Q.
Ans. P=q(2a)
Unit =Cm
Q.
Ans. 10V
2012
Q.Why should electrostatic field be zero inside a conductor?
Ans If the electric field inside the conductor is not zero, the electrons will accelerate
due to the electric field and for the electrostatic condition the net field become zero
due to the redistribution of the charge carries and electrons come at rest
(electrostatics).
Q
Ans
Ans.
2009
Q
Define electric flux. Write its S I units.A change q is enclosed by a spherical surface of
radius R. If the radiusis reduced to half how would the electric flux through the surface
change?
Ans.Electric flux is the number of field lines crossing an area. It is
given byⱷ= E.dS. Its SI unit is N.m2/C.
The electric flux through a spherical surface of radius R for a
charge q enclosed by the surface is q/εo. If radius is reduced to half,
the electric flux remains the same.
Q.Can two equipotential surfaces intersect each other? Give reasons.
Two charges –q and +q are located at points A(0,0,-a) and B(0,0,+a)
respectively.What is the work done in moving a test charge from point P (7,0,0) to Q
(-3,0,0)?
2
Ans.No
2010
Q.
2011
Q.
Q.
Q
Q.
Q.
2009
State Gauss’s law in electrostatics.Using this law derive an expression for the
electric field intensity at a near point due to an infinite thin plane sheet of uniform
charge density. 3
2010
Q.
Ans.i) Capacitance of the capacitor increases by a factor K ,i.e, it becomes KC.
(ii) Net electric field will get reduced. As potential difference V =-Ed, as E is reduced,
potential difference between the capacitor plates also reduces.
(iii) Energy of the capacitor:
As the charge Q is fixed on plates,
Energy stored in the capacitor
x (Energy without dielectric).
So,
it goes down.
Q.
Ans.
(a) An equipotential surface is a surface with a constant value of potential
at all points on the surface.
Equipotential surfaces for two identical positive charges.
(b) First, we calculate the work done in bringing the charge q1 from infinity to r1.
Work done in this step is q1 V(r1).
Next, we consider the work done in bringing q2 to r2. In this step, work is done not
only against the external field E but also against the field due to q1.
Work done on q2 against the external field = q2V (r2)
Work done on q2 against the field due to q1 =
wherer12 is the distance between q1 and q2. By the superposition principle for fields,
we add up the work done on q2 against the two fields (E and that due to q1):
Work done in bringing q2 to r2 =
Thus,
Potential energy of the system
= the total work done in assembling the configuration
2011
Ans
Ans
Consider a spherical Gaussian surface of radius r (›R), concentric with given shell. If
is electric field outside the shell, then by symmetry, electric field strength has
same magnitude
on the Gaussian surface and is directed radially outward. Also
the directions of normal at each point is radially outward, so angle between
and
is zero at each point. Hence, electric flux through Gaussian surface =
Now, Gaussian surface is outside the given charged shell, so charge enclosed by
the Gaussian surface is Q.
Hence, by Gauss's theorem
Thus, electric field outside a charged thin spherical shell is same as if the whole
charge Q is concentrated at the centre. Graphically,
For r ‹ R, there is no strength of electric field inside a charged spherical shell.
For r › R, electric field outside a charged thin spherical shell is same as if the whole
charge Q is concentrated at the centre.
2013
Q.
A capacitor of unknown capacitance is connected across a battery of V volts. The
charge stored in it is 360 µC. when potential across the capacitor is reduced by 120
V, the charge stored in it becomes 120µC. Calculate:
(i) The potential V and unknown capacitance C.
(ii) What will be the charge stored in the capacitor, if the voltage applied had
increased by 120 V?
3
Q.
A hollow cylindrical box of length 1m and area of cross section 25cm 2 is placed in a
three dimensional coordinate system( whose axis along X- axis and nearer cross

section at 1m from the origin). The electric field in the region is given by E = 50x iˆ ,
where E is in NC-1 and x is in metres. Find
(i) Net flux through the cylinder.
(ii) Charge enclosed by the cylinder.
3
Ans.
If voltage applied have increased by 100 V:
Charge stored will be=
Or
2010
Ans
2012
Ans.
5