Download MODULE :2 Lecture 6 Multiple Choice Questions : 1. Eight

MODULE :2
Lecture 6
Multiple Choice Questions :
1. Eight charges, each +q, are located symmetrically on a circle of radius R with P as its
centre. If the charge at the position X is removed and brought to the location P, the
force on this charge will be
a.
b.
c.
d.
1 𝑞𝑞 2
from X to P
4𝜋𝜋𝜋𝜋 𝑅𝑅 2
7 𝑞𝑞 2
from X to P
4𝜋𝜋𝜋𝜋 𝑅𝑅 2
1 𝑞𝑞 2
4𝜋𝜋𝜋𝜋 𝑅𝑅 2
7 𝑞𝑞 2
4𝜋𝜋𝜋𝜋 𝑅𝑅 2
X
from P to X
P
from P to X
2. Twelve charges are positioned on the dial of a wall clock such that a charge Q is at the
position 1, 2Q at position 2, 3Q at 3 and so on, ending with 12Q at the position 12. If O is
at the centre of the dial, what is the force exerted on a unit charge located at O,
assuming the radius of the dial to be of unit length?
y
a. �12 + 6√3�𝑖𝑖̂ + 6𝑗𝑗̂
b. −�12 + 6√3�𝑖𝑖̂ + 6𝑗𝑗̂
c. �12 + 6√3�𝑖𝑖̂ − 6𝑗𝑗̂
d. −�12 + 6√3�𝑖𝑖̂ − 6𝑗𝑗̂
x
3. Two small conducting spheres attract each other electrostatically. It can be concluded
that
a. At least one of the spheres is charged
b. Both the spheres are charged
c. Both the spheres are charged and their charges are of opposite sign
4.
5.
6.
7.
8.
d. No definite conclusion on their charge state can be made from the given data.
Electric field lines are
a. Vectors in the direction of the electric force that acts on a test charge
b. Trajectories of a test charge in the electric field
c. Closed loops
d. Pictorial representation of electric field around a charged object.
The total amount of negative charge of all the electrons contained in one mole of water
is approximately
a. 1C
b. 105 C
c. 106C
d. 1023C
A negative charge of 9µC and mass 2µkg orbits around a heavy positive charge of 16µC
in a circular orbit of radius 5m. What is the speed of the negative charge?
a. 180 m/s
b. 324 m/s
c. 360 m/s
d. 1984 m/s
A 2.5 µC test charge is placed to the right of another charge Q. If there is an attractive
force of 22.5 N between the two, what would be the force exerted if the magnitude of
the test charge were to be doubled but it stayed at the same location as before?
a. 90 N
b. 45 N
c. 11.25 N
d. Depends on the distance between the test charge and the charge Q.
The charges 𝑄𝑄, 𝑄𝑄, −4𝑄𝑄and − 𝑄𝑄 are kept at the corners A, B, C and D , respectively, of a
square. If the force between the charges at A and B is F, the net force exerted on the
charge at A due to the other three charges is
a.
b.
c.
d.
e.
6F
3𝐹𝐹
√6𝐹𝐹
√3𝐹𝐹
A
B
D
C
9. Two small conducting spheres of mass m and charge q each are suspended from a
common point by means of threads. The spheres settle down to an equilibrium position,
each making an angle θ with the vertical. The tension in either of the threads is
a.
b.
c.
d.
Zero
Greater than mg
Less than mg
Equal to mg
10. Two opposite charges are placed on the paper, as shown in the figure.
The charge on the left is three times as big as the charge on the right. Other than at infinity,
where else can the force on a unit test charge due to these two charges is zero?
d
a.
b.
c.
d.
To the right of the smaller charge
Between the two charges
To the left of the bigger charge
Depends on the sign of the test charge
Answers to Multiple Choice Questions
1. (b) 2. (c) 3. (a) 4. (d) 5. (c) 6. (c) 7. (b) 8. (c) 9. (b) 10. (a)
Problems :
1. ABC is an equilateral triangle of side 40 cm. At the vertices A and B +4µC of charge is
kept fixed while at the vertex C a −4𝜇𝜇C charge is held. What is the force on the charge
at C? If now, the charge at C is released while the charges at A and B still remain fixed,
describe its subsequent motion.
A
C
B
2. Two equal and positive charges, q each, are at a finite distance 2d from each other. A
third charge Q is located at the midpoint of the line joining the two. Where should a unit
positive charge be placed so that the net force on it zero? What, if any, is the
requirement of the magnitude and sign of the charge Q?
3. Twelve equal charges +q are situated on a circle of radius R and they are equally spaced
like the position of digits on the dial of a clock. What is the net force on a charge Q kept
at the centre? What would be the force on Q if the charge at 3’o clock position is
removed?
4. Two particles, each of mass m and having charges q and 2q are suspended by strings of
length l from a common point. Find the angle θ that each string makes with the vertical.
5. A particle of mass m and with charge q is suspended from a peg on a wall by means of
a string of length 0.5m. The string makes an angle 600 with the vertical. Another charge
q is held at the same horizontal level as the first charge so that the distance between
the charges is R. Calculate the tension in the string and the distance R.
6. A wire is bent in the form of a semicircle and carries a linear charge density λ. Find the
electric field at the centre of the circle.
Hints for solutions to problems :
1. Resolve the forces along the perpendicular bisector from C on to AB and a direction
perpendicular to it. The net force is along the perpendicular bisector towards AB and
has a magnitude 1.56 N.
2. Charge Q has to have opposite sign of the charge q; else at no finite distance field can be
zero. In order that forces cancel out, the test charge must be located along the
perpendicular at Q of the line joining the two charges q.
q
Q
q
θ
P
Resolve the forces due to the pair q, the components perpendicular to QP cancel. Along
QP, it has to be canceled by the repulsive force due to Q. The condition gives cos3 𝜃𝜃 =
𝑄𝑄/2𝑞𝑞.
3. When equal charges are kept on the dial, the net force on a charge at the centre is zero
as the forces due to a pair kept in diametrically opposite positions cancel. If a single
charge is removed, the effect is due to the diametrically opposite member.
4. Draw the freebody diagram of each particle. Resole forces along vertical and horizontal.
Fe
𝑻𝑻 𝐜𝐜𝐜𝐜𝐜𝐜 𝜽𝜽 = 𝒎𝒎𝒎𝒎
𝑻𝑻 𝐬𝐬𝐬𝐬𝐬𝐬 𝜽𝜽 =
𝟏𝟏
θ
T
2q
q
mg
𝟐𝟐𝒒𝒒𝟐𝟐
𝟒𝟒𝟒𝟒𝝐𝝐𝟎𝟎 𝟒𝟒𝒍𝒍𝟐𝟐 𝐭𝐭𝐭𝐭𝐭𝐭𝟐𝟐 𝜽𝜽
Eliminate T to get the angle θ.
5. Very similar to Problem 4. The right hand side of the second equation is to be changed
to
𝟏𝟏
𝒒𝒒𝟐𝟐
𝟒𝟒𝟒𝟒𝝐𝝐𝟎𝟎 𝑹𝑹𝟐𝟐
. Since the angle is given and T can be eliminated, one can find R.
6. Consider the field at P due to an element Rdθ located at an angle θ, as shown. The
magnitude of the field due to this element is given by 𝑑𝑑𝑑𝑑 =
1
4𝜋𝜋𝜖𝜖 0
𝜆𝜆𝜆𝜆 𝑑𝑑𝑑𝑑
𝑅𝑅 2
. The component
of the field parallel to the diameter cancels due to a symmetrically placed element,
leaving the net field to be perpendicular to the diameter and pointing away from the
semicircle.
θ
P
The net field has a magnitude |𝐸𝐸| =
1
4𝜋𝜋𝜖𝜖 0
𝜋𝜋
2 ∫02
𝜆𝜆𝜆𝜆 𝑑𝑑𝑑𝑑
𝑅𝑅 2
sin 𝜃𝜃 =
𝜆𝜆
2𝜋𝜋𝜖𝜖 0 𝑅𝑅