Case Study 1: Coulomb's Law
Two point charges q1 and
q2 are placed at a certain distance rrr apart in a vacuum. The force of
interaction between these charges is found to be proportional to the product of
their magnitudes and inversely proportional to the square of the distance
between them. This force is described by Coulomb’s law, which is expressed
mathematically as:
F=\(\ \frac{k q_1
q_2}{r^2}\)
where k is Coulomb’s
constant. This force is attractive if the charges are of opposite signs and
repulsive if the charges are like (both positive or both negative).
Questions:
-
According to Coulomb’s law, the force between two
charges is:
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a) Directly proportional to the distance between them
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b) Inversely proportional to the product of the charges
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c) Inversely proportional to the square of the distance
between them
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d) Independent of the medium between the charges
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If the distance between two charges is halved, the
force between them will:
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a) Double
-
b) Quadruple
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c) Remain the same
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d) Become half
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Coulomb’s law is similar to which of the following
laws?
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a) Hooke’s law
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b) Newton’s law of gravitation
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c) Ohm’s law
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d) Faraday’s law
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What happens to the force between two charges if they
are placed in a medium with a dielectric constant ε?
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a) It increases
-
b) It decreases
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c) It remains unchanged
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d) It becomes zero
Answers:
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c) Inversely proportional to the square of the
distance between them
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b) Quadruple
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b) Newton’s law of gravitation
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b) It decreases
Case Study 2: Electric Field Due to a Point Charge
A point charge qqq creates an
electric field around it, which exerts a force on any other charge placed in the
vicinity. The strength of the electric field EEE at a distance rrr from the
point charge is given by:
E=\(\ \frac{k
q}{r^2}\)
where k is Coulomb’s
constant. The direction of the electric field is radial, pointing away from the
charge if q is positive and toward the charge if qqq is negative.
Questions:
-
The electric field due to a point charge varies with
distance as:
-
a) \(\frac{1}{r}\)
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b) \(\frac{1}{r^2}\)
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c) \(\ {r^2}\)
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d) r
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If a positive charge is placed in the electric field
of another positive charge, the force experienced by the placed charge will be:
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a) Attractive
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b) Repulsive
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c) Zero
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d) Random
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What is the direction of the electric field created by
a negative point charge?
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a) Away from the charge
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b) Toward the charge
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c) Perpendicular to the charge
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d) Tangential to the charge
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If the distance between the charge and the point where
the electric field is measured is doubled, the electric field strength:
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a) Becomes half
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b) Becomes one-fourth
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c) Becomes double
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d) Remains the same
Answers:
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b) \(\ \frac{1}{r^2}\)
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b) Repulsive
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b) Toward the charge
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b) Becomes one-fourth
Case Study 3: Electric Dipole in a Uniform Electric Field
An electric dipole consists of
two equal and opposite charges separated by a small distance. When placed in a
uniform electric field, the dipole experiences no net force, but it experiences
a torque that tends to align the dipole with the field. The torque \(\tau \) acting
on the dipole is given by:
\(\tau = pE \sin \theta \)
where p is the dipole moment,
E is the electric field strength, and θ\thetaθ is the angle between the dipole
moment and the electric field.
Questions:
-
An electric dipole in a uniform electric field
experiences:
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a) A net force
-
b) No force but a torque
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c) Both a force and a torque
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d) Neither a force nor a torque
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The torque acting on a dipole in an electric field is
maximum when the angle between the dipole and the field is:
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a) \(\ 0^\circ \)
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b) \(\ 45^\circ \)
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c) \(\ 90^\circ \)
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d) \(\ 180^\circ \)
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What happens to the dipole in a uniform electric field
if it is initially aligned with the field?
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a) It experiences maximum torque
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b) It experiences no torque
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c) It rotates and aligns perpendicular to the field
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d) It gains potential energy
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The dipole moment is a vector quantity directed:
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a) From the positive to the negative charge
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b) From the negative to the positive charge
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c) Along the axis of the dipole
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d) Perpendicular to the axis of the dipole
Answers:
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b) No force but a torque
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c) \(\ 90^\circ \)
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b) It experiences no torque
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a) From the positive to the negative charge
Case Study 4: Gauss’s Law
Gauss’s law relates the
electric flux through a closed surface to the charge enclosed within that
surface. It is mathematically expressed as:
\(\ \Phi_E =
\frac{q_{\text{enc}}}{\varepsilon_0} \)
where \(\ \Phi_E \) is the
electric flux, \(\ q_{\text{enc}} \) is the enclosed charge, and
\(\ \varepsilon_0 \) is the permittivity of free space. This law simplifies the
calculation of electric fields in situations with high symmetry.
Questions:
-
Gauss’s law relates the electric flux through a closed
surface to:
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a) The total charge outside the surface
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b) The total charge inside the surface
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c) The electric field outside the surface
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d) The magnetic flux through the surface
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According to Gauss’s law, the electric field inside a
hollow conducting shell is:
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a) Zero
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b) Constant
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c) Equal to the external field
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d) Infinite
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Gauss’s law is particularly useful in calculating the
electric field for which of the following geometries?
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a) Asymmetric charge distributions
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b) Spherical charge distributions
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c) Arbitrary charge distributions
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d) Complex charge distributions
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In Gauss’s law, the electric flux through a closed
surface depends on:
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a) The shape of the surface
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b) The size of the surface
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c) The charge enclosed within the surface
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d) The position of the charge outside the surface
Answers:
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b) The total charge inside the surface
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a) Zero
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b) Spherical charge distributions
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c) The charge enclosed within the surface
Case Study 5: Electric Field Lines
Electric field lines are a
visual representation of electric fields. They show the direction of the force
on a positive test charge placed in the field. The density of the field lines
indicates the strength of the electric field, and the lines never cross each
other. Field lines emerge from positive charges and terminate on negative
charges.
Questions:
-
Electric field lines provide information about:
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a) The direction of the force on a negative charge
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b) The direction of the force on a positive charge
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c) The magnitude of the charge
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d) The potential at different points in space
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Electric field lines around a positive point charge:
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a) Emerge radially outward
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b) Emerge radially inward
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c) Are circular
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d) Are perpendicular to the charge
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The strength of the electric field at a point is
represented by:
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a) The number of field lines passing through the point
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b) The color of the field lines
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c) The length of the field lines
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d) The shape of the field lines
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Electric field lines never cross each other because:
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a) They represent different fields
-
b) Two directions of force at one point would be
contradictory
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c) The field is zero at the crossing point
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d) The charges cancel out at the crossing point
Answers:
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b) The direction of the force on a positive charge
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a) Emerge radially outward
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a) The number of field lines passing through the point
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b) Two directions of force at one point would be
contradictory