Hot Questions
UNIT 01: ELECTROSTATICS
1 Marks Questions:
Q.1 A certain region has cylindrical symmetry of electric field. Name
the charge distribution producing such a field.
Q.2 Represent graphically the variation of electric field with
distance, for a uniformly charged plane sheet.
Q.3 How will the radius of a flexible ring change if it is given
positive charge?
Q.4 Five Charges of equal amount (q) are placed at five corners of a
regular hexagon of side 10 cm. What will be the value of sixth charge placed at
sixth corner of the hexagon so that the electric field at the centre of hexagon
is zero ?.
Q.5 Two conducting spheres of radii r1 & r2
are at same potential. What is the ratio of charges on the spheres?.
Q.6 Why do we use nitrogen or methane gas in Van-de-Graff generator ?
Q.7 An electric charge q is placed at one of the corner of a cube of
side ‘a’. What will be the electric flux through its one of the face?
Q.8 A square surface of side L meters is in the plane of the paper. A
uniform electric field E (volts/m), also in the plane of paper, is limited only
to lower half of the square . What will be the electric flux (in SI units) associated with the surface.
Q.9 The distance of the field point on the equatorial plane of a small
electric dipole is halved. By
what factor will the electric field change for the dipole?
2 Marks Questions:
Q.10 A charged particle is free to move in an
electric field. Will it always move along an electric line of force? Justify
your answers?.
(2 marks)
Q.11 A charge Q is divided in
two parts q and Q - q separated by a distance R. If force between the two
charges is maximum, find the relationship between q & Q.
(2 marks)
3 marks question:
Q12. A parallel plate
capacitor is charged to potential
V by a source of emf . After
removing the source, the separation
between the plates is doubled . How will the following change electric field change
on each plate potential difference capacitance of the capacitor Justify your
answer
Q 13 If N
drops of same size ,each having the same charge ,coalesce to form a bigger drop
. How will the following vary with respect to single small drop?
(i)Total charge on bigger drop
(ii) Potential on the bigger drop
(iii) The capacitance on the bigger drop
Q14 Work done to move a
charge along a closed path inside an electric field is always zero, using this
fact, prove that it is impossible to produce an electric field in which all
limes of force would be parallel lines and density of their distribution would
constantly increase in a direction perpendicular to the lines of force.
Q 20. Obtain the formula for
electric field due to a long thin wire of uniform linear charge density without using Gauss’s law.
ANSWERS/HINTS
1 mark question
Q.1 Uniform linear charge distribution
Q.2 E is constant with r.
Q3. Increases due to repulsion
Q.4 6th Charge is Q
Q.5 Q1/Q2=R1/R2
Q.6 It transfers the leakage of Charge to earth through earthed steel chamber
Q.7 Q/24 έ0
Q.8 Zero
Q9. Ex=Ez<Ey
Q.10 E a 1/r3 if
r=r/2 , E= 8 times
2 marks question
Q 11 work done is independent of path
w= ¼πεΟ q1 q2 ( 1/r1-
1/r2) putting the values
& ans 15 J
Q 12 if charge is positive
& at rest in electric field then it will move along electric line of force.
If charge has initial velocity making some angle with electric field than it
will along parabolic path.
Q 13 Ф'= 4Ф
Q + q1+ q2+q3/
εo= 4 X (q1+q2+q3)/εo
putting the values & finding Q = 3*8.854
μC
Q 14. F=K q(Q-q) / r 2
for max. &min.dF / dq=0 , q=Q/2
Q 15. All are in parallel
C= εoA/3d+εoA/6d+
εoA/9d= 11εoA/18 d
3 marks question
Q.16 a. E same
b. Q same
c. V same
d. C is halved with reasons
Q.17 i. N times the charge
on small drop
ii. N 2/3 times the
potential on small drop
iii N 1/3 times
the capacitance on small drop
Q.18 If q is moved along abcd then Wabcda = 0
Wab+ Wbc+ Wcd +Wda
= 0
as E perpendicular to bc & da
so Wbc = Wda = 0
therefore Wab= -Wcd
But Wab can never be equal to Wcd as
the lines of force are closer to cd
therefore Wcd > Wab
therefore Wabcda is not equal to 0 hence such
electric field E is impossible
Q.19 i As the charge move
closer the charge on large sphere ` is
redistributed as shown in diagram
ii As the
spheres move more closer than the charge is redistributed as shown in diagram
iii Behaviour of
force between 2 cm & 1 cm :
force of repulsion increases upto 1.4 cm
& F rep. is max.
at r=1.4 cm
If 1.2 cm<r<1.4 cm F rep. is
decreasing Fatt. increases due to inductive effect.
At r=1.2 cm F rep. = F att. & if
r<1.2 cm force is strongly attractive
Q.20 X=q/length = q/l change on dl length
dq = λ dl At point l
dE =
1 / 4 π εo * dq /op2 = 1/ 4π εo * λ dl / r2 +l2 l= rtan Q
find dl .dE = 1 / 4 π εo λ cos q dq/ r
integrate between -π/2 to +π /2 than E is λ/2π εo
r
thanks sir....for providing such a useful study material....thanks a lot sir!
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