test paper 12 cbse

1. Roll No. :
Date :
Time -
MM - 50
1
1.
Ans :
1
2.
Ans :
1
3.
Ans :
2
4.
The given figure shows a non-conducting semicircular rod. What is the direction of the net
electric field at point P due to the charge on the rod?
Electric field will be along –y direction because x-axis components due to upper and
lower halves of the semicircular charged rod will get cancelled out.
A uniform electric field E exists between two charged plates as shown in figure. What would be
the work done in moving a charge q along the closed rectangular path ABCDA?
Work done in moving a charge q along a closed rectangular path ABCD is calculated as
W = WAB + WBC + WCD + WDA
W = qE + 0 – qE + 0 = 0 [ AB = CD]
The following graph shows the variation of charge Q, with voltage V, for two capacitors K and L.
In which capacitor is more electrostatic energy stored?
In capacitor L, more electrostatic energy is stored.
Figure shows two large metal plates, P1 and P2, tightly held against each other and placed
between two equal and unlike point charges perpendicular to the line joining them.
(i) What will happen to the plates when they are released?
(ii) Draw the pattern of the electric field lines for the system.

2. Ans :
2
5.
Ans :
2
6.
Ans :
2
7.
A thin straight infinitely long conducting wire having charge density λ is enclosed by a cylindrical
surface of radius r and length l, its axis coinciding with the length of the wire. Find the expression
for the electric flux through the surface of the cylinder.
According to the Gauss’s law, the electric flux through a closed surface is times the
charge enclosed by the surface.
As the charge enclosed by the cylindrical surface is q = λl.
The figure given below shows a uniformly charged non-conducting rod. What is the direction of
electric field at point P due to the charge on the rod?
From the figure, we see that x-axis components of electric field due to upper and lower
halves of the rod will get cancelled out. Therefore, net electric field will be in – y-axis.
A small metal sphere carrying the charge +Q is located at the centre of a spherical cavity in a
large uncharged metal sphere

3. Ans :
2
8.
Ans :
as shown in the figure.
Use the Gauss’s theorem to find the electric flux at points P1 and P2.
Let us draw a Gaussian sphere of radius r, passing through point P1, then net electric
flux through the sphere
Now, we draw another Gaussian sphere of radius r2 passing through point P2.
As we can see, –Q charge will be induced on the inner side of the cavity of metal sphere.
Net electric charge enclosed = Q – Q = 0
S1 and S2 are two hollow concentric spheres enclosing charge Q and 2Q respectively as shown
in figure.
(i) What is the ratio of the electric flux through S1 and S2?
(ii) How will the electric flux through the sphere S1 change, if a medium of dielectric constant 5 is
introduced in the space inside S1 in place of air?
(i) According to the Gauss’s law, electric flux (Φ) is given by
When a medium of dielectric constant K = 5 is introduced inside S1, then the electric flux
through S1
i.e. the flux will be reduced to th of its initial value.

4. 2
9.
Ans :
3
10.
Ans :
3
11.
A graph is drawn between some physical quantity x and r as shown below, where r is the
distance from the centre of a charged conducting sphere.
Now answer the following:
(a) Name the physical quantity x.
(b) At what point electric field is (i) maximum, and (ii) minimum?
(a) x is an electric potential, i.e. V on the charged conducting sphere.
(b) (i) Electric field is maximum at B , and (ii) minimum at A (zero).
A hemispherical surface lies as shown in an uniform electric field region. Find the net electric flux
through the curved surface if electric field is
(a) along x-axis, and
(b) along y-axis.
(a) Since, the number of field lines entering the hemisphere is equal to number of field
lines leaving. Hence, the net electric flux through it is zero.
(b) As no charge is enclosed, therefore net electric flux is given by
State Gauss’s law in electrostatics. A cube with each side a is kept in an electric field given by
, (as is shown in the figure) where C is
a positive dimensional constant. Find out
(i) the electric flux through the cube, and
(ii) the net charge inside the cube
.

5. Ans :
3
12.
Ans :
3
13.
Ans :
3
14.
Gauss’s law : The surface integral of electric field (electric flux) over any closed surface
is times the charge enclosed in it.
An uncharged comb after combing hair, when brought near the paper bits attracts them. Answer
the following:
(a) Does the mass of comb/paper bit get changed?
(b) Is paper bit still uncharged?
(c) What is the difference between the charging of a comb and the charging of the paper bits?
(a) Yes, by negligible amount.
(b) Yes.
(c) The charging of comb is due to charging by friction.
The charging of paper bits is due to charging by induction.
A thin metallic spherical shell of radius R carries a charge Q on its surface. A point charge is
placed at its centre
C and an other charge +2Q is placed outside the shell at a distance x from the centre as shown in
figure. Find (i) the force on the charge at the centre of shell and at the point A, (ii) the electric flux
through the shell.
(i) As there is no electric field inside thin charged metallic spherical shell, the force on the
charge is zero.
Force on charge 2Q at point A.
(ii) Electric flux through the shell,
Three concentric metallic shells A, B and C of radii a, b and c (a b c) have surface charge
densities + σ, – σ and + σ respectively as shown in the figure.
If shells A and C are at the same potential, then obtain the relation between the radii a, b, c.

6. Ans :
2
15.
Given that, A, B and C are three concentric shells of radii a, b and c. σ, –σ and σ are the
charge densities on them respectively.
Also VC = VA (given)
Five identical horizontal square metal plates each of area A are placed at a distance d apart in
air and connected to the terminals A and B as shown in the figures (a) and (b). Find the effective
capacitance between the two terminals A and B.

7. Ans :
3
16.
Ans :
5
17.
Figure shows three circuits, each consisting of a switch and two capacitors initially charged as
indicated. After the switch has been closed, in which circuit (if any) will the charges on the left hand
capacitor (i) increase, (ii) decrease and (iii) remain same?
(i) Before closing the switch,
Charge on left hand conductor remains same.
(ii) When s is open,
Therefore, the charge on left hand conductor decreases.
(a) Two isolated metal spheres A and B have radii R and 2R respectively, and same charge q.
Find which of the two spheres have greater:
(i) capacitance and (ii) energy density just outside the surface of the spheres.
(b) (i) Show that the equipotential surfaces are closed together in the regions of strong field and far
apart in the regions of weak field. Draw equipotential surfaces for an electric dipole.

8. Ans :
5
18.
(ii) Concentric equipotential surfaces due to a charged body placed at the centre are shown.
Identify the polarity of the charge and draw the electric field lines due to it.
(a) Given: RA = R, RB = 2R
(i) Capacitance, C = 4πε0R
i.e. C ∝ R
Hence, the capacitance of sphere B is more.
Therefore, sphere A has greater energy density.
Here ∆r = separation between two equipotential surfaces
Therefore, ∆r will be less, if E will be more, i.e. equipotential surfaces are closer if electric
field is stronger and vice versa.
(ii) In the direction of electric field, the electric potential decreases. Therefore, an electric
charge at the centre is negative.
(a) Deduce the expression for the potential energy of a system of two charges q1 and q2 located
at and respectively in an external electric field.
(b) Three point charges, +Q, + 2Q and –3Q are placed at the vertices of an equilateral triangle ABC
of side l. If these charges are displaced to the mid-points A1, B1 and C1 respectively, find the
amount of the work done in shifting the charges to the new locations.

9. Ans :
5
19.
(a) Refer to Ans. 61(b).
(b) Electrostatic potential energy of the systems of charges corresponding to initial
configuration is
Electrostatic potential energy of the system of charges corresponding to final
configuration is
The amount of work done in shifting the charges to new locations is
Derive the expression for the energy stored in a parallel plate capacitor of capacitance C with air
as medium between its plates having charges Q and where A is the area of each plate
and d is the separation between the plates.
How will the energy stored in a fully charged capacitor change when the separation between the
plates is doubled and a dielectric medium of dielectric constant 4 is introduced between the plates?

10. Ans : Suppose at any instant of time potential difference between the capacitor plates be V.
Then the amount of work required to supply a charge dq to the capacitor is given by
dW = Vdq
To supply a charge Q, the work done is given by
This work done gets stored in the form of electrostatic potential energy
On introducing the dielectric medium of dielectric constant K = 4 between the plates,
The energy stored in the capacitor reduces to one-half of its original value.