1. ELECTRIC CHARGE (Chapter 22)
Key concepts:
Charged particles attract to or repel from each other. If particles are of the same charge
the repel from each other, if they are of the opposite charge they attract to each other.
Units of charge: Coulombs.
The Coulomb law states that the value of the force between two charged particles Q1 and
Q2 is proportional to their charges and reciprocally proportional to the square of the
distance r between them:
QQ
F = k 12 2
r
9
2
2
where coefficient k = 8.99 × 10 Nm / C which is sometimes expressed via another
1
ε0 =
= 8.85 × 10− 12 C 2 / Nm 2
4π k
constant
Concepts of charge densities: for charge Q uniformly distributed among some volume V,
one can introduce a charge per unit volume or volume charge density which is the ratio:
ρ = Q /V
If object is flat, or, in general, 2 dimensional one can introduce the surface charge density
which is the ratio between the charge Q distributed among the surface area S: σ = Q / S .
For linear, or 1 dimensional objects, the linear charge density can be considered, i.e
charge per unit length: λ = Q / L . We therefore have:
3D : ρ = Q / V
2D : σ = Q / S
1D : λ = Q / L
Typical problems related to electric charge:
Problem 1. Two electrons are placed at a certain distance between each other? Force
between them:
A. Repulsion
B. Attraction
C. Zero
D. Not determined
E. None of the above
Solution.
Two electrons have negative charge, therefore the force is repulsion.
Answer A
Problem 2. Find force between proton and electron placed at the distance 1 µm.
A.
B.
C.
D.
E.
2.3 * 10-13N, attraction
2.3 * 10-13N, repulsion
2.3 * 10-16N, attraction
2.3 * 10-16N, repulsion
2.3 * 10-9N, attraction
Solution.
Force between two oppositely charged particles is attraction and its value is
−19
q1q2
× 1.6 × 10 −19
9 1.6 × 10
8.99
10
=
×
= 2.3 × 10− 16
2
−6 2
r
(10 )
Answer C.
F =k
Problem 3. A postively charged particle with Q= 5 mC is placed between two negatively
charged particles with q1= 1 mC (left) and q 2=9 mC (right). The distance between q 1 and
q is 5 cm and the distance from q to q 2 is 9 cm. What is the total force acting on the
middle particle? Find the value and the direction.
A) 3.2 x 10 7 N, left
B) 3.2 x 10 7 N, right
C) 2.8 x10-9 N, left
D) 2.8 x 10 -9 N, right
E) Zero
Solution.
The value of the force from particle 1 onto the middle particle is
F1 = k
−3
−3
q1Q
9 1 × 10 × 5 × 10
=
8.99
×
10
= 1.8 × 107
2
−2 2
r1
(5 × 10 )
The direction of this force is to the left since particle 1 and the middle particle are
oppositely charged.
The value of the force from particle 2 onto the middle particle is
F2 = k
−3
−3
q2Q
9 9 × 10 × 5 × 10
8.99
10
=
×
= 5.0 × 107
2
(9 × 10− 2 ) 2
r2
The direction of this force is to the right since particle 2 and the middle particle are
oppositely charged.
If we choose x axis from the left to the right, the force F1 is projected with minus sign
(since its direction is opposite to the x-axis) and the force F2 is projected with plus sign
(since its direction is along the x-axis). The total force is therefore
F x tot = F1 + F2 = −1.8 × 10 7 + 5.0 ×10 7 = 3.2 × 10 7 N
since it is positive, the total force is pointed to the right.
x
x
Answer B.
Problem 4. A negatively charged particle with q= -1 µC is placed at the center of the
uniformly charged cube with side a=1 mm and total charge Q=3 C. What is the total force
acting on the particle in the center? Find the value and the direction.
A) 0.75 x 10 6 N, left
B) 0.75 x 10 6 N, right
C) 2.8 x10-9 N, top
D) 2.8 x 10 -9 N, bottom
E) Zero
Solution. Since the cube is uniformly charged and it has a central symmetry. The force
from any infinitesimal charge located at the point r will be compensated by the force
from the same infinitesimal charge located at the point –r. Therefore, the total force
acting on any charge placed at the center is equal zero by symmetry.
Answer E.
Problem 5. What is the surface charge density of the sphere with R=1 cm and Q=1 µC
uniformly distributed among the sphere surface:
A) 0.01 C/m2
B) 124 C/m2
C) 0.45 x 10-7 C/m2
D) 7.96 C/m2
E) 0.0034 C/m2
Solution.
The surface charge density is the charge per unit area of the sphere surface. It is
therefore:
σ = Q / S = Q / 4π R 2 = 1 × 10 −6 /[4*3.14*(1 × 10 −2 )2 ] = 7.96C / m 2
Answer D.
2. ELECTRIC FIELDS (Chapter 23)
Key concepts:
A charged particle creates electric field around it. If the particle is positively charged, the
electric field is pointed outwards, if the particle is negatively charged, the electric field is
pointed inward.
The value of the electric field at some point r from the charged particle located at the
origin is given by
q
E( r ) = k 2
r
Note that the electric field is a vector field, i.e. at any point of space it has the value and
the direction, so we can write a more general formula:
r
r r
q r
E( r ) = k 2
r r
r
r
where the vector r is a unit vector pointing in the direction of the field.
If another (test) particle is placed inside the electric field, it will experience the Coulomb
force:
r
r
F = q testE
Note that the direction of this force depends on the direction of the field and the sign of
the charge of the test particle: if the test particle is positively charged, it experiences the
force along the field, if the test particle is negatively charged its force is opposite to the
field direction. Note also that the particle will accelerate in the electric field since
according to the Newton second law:
r
r
r
F = q test E = ma
Units of electric field: N/C
Useful formulae for electric fields created by various objects:
A uniformly charged infinite plate creates an electric field, which is a constant in all
σ
E=
2ε 0 where σ is the surface charge density on the plate.
space, and it is given by
A uniformly charged line with linear density λ creates an electric field, which is given
λ
E=
2πε 0 R
by:
Typical problems related to electric fields:
Problem 6. A negatively charged particle is placed in a uniform electric field directed
from South to North. In which direction the particle will move after it is released?
A) West
B) East
C) South
D) North
E) North-West
Solution. A negatively charged particle placed into the electric field will move in the
direction opposite to the field. Therefore, it is South.
Answer C.
Problem 7. A negatively charged particle accelerates from East to West inside a uniform
electric field. What are the direction and the value of the electric field if the particle has
charge q=3 µC, mass m=1 mg, and the value of its acceleration is 3 mm/s 2? Select the
closest answer:
A) West, 0.001 N/C
B) East, 0.001 N/C
C) West, 1 N/C
D) East, 1 N/C
E) North, 1000 N/C
W
E
Solution. If the particle is moving from East to West and it is negatively charged, the
electric field is opposite to the particle movement, therefore it is directed to the West.
Its value can be found from the second Newton law: qE = ma and it is given by
a = ma / q = 10 −6 × 3 × 10 −3 / 3 × 10− 6 = 0.001
Answer A.
Problem 8. Consider a dipole placed in a constant electric field. The field is directed
from the top to the bottom. The dipole is initially positioned with the positive charge on
the left and the negative charge on the right. What happens after we release the dipole?
A) The dipole will be rotated by 90 degrees, positive charge
is on the top, negative charge is on the bottom.
B) The dipole will be rotated by 90 degrees, positive charge
is on the bottom, negative charge is on the top.
C) The dipole will be rotated by 180 degrees, positive charge
is on the right, negative charge is on the left.
D) The dipole will not be rotated, positive charge remains on the left, negative charge
remains on the right.
E) The dipole will be rotated by 45 degrees, positive charge is on the top-left, negative
charge is on the bottom-right.
Solution. The negative charge of the dipole will move in the direction opposite to the
field while the positive charge will move along the field. Remember that the distance
between negative and positive charge in the dipole remains fixed. Therefore, the dipole
will be rotated by 90 degrees with negative charge on the top and positive charge on the
bottom. After that, the dipole will not move since the force on the negative charge will
completely compenstate the force on the positive charge and the total force on the dipole
is equal zero.
Answer B.
Problem 9. A small particle with mass m=1 mg and positive charge Q=1 mC is placed
just near the ground. What should be the surface charge density on the ground to keep the
particle above it?
A) 1.75 x 10 -19 C/m2
B) 1.75 x 10-16 C/m2
C) 1.75 x 10-13 C/m2
D) 1.75 x 10 -10 C/m2
E) 1.75 C/m2
Solution. Since the particle is small comparing to the ground, we can assume the ground
to be flat infinite plate. Distributing the positive charge on the plate will create the
electric field E = σ / 2ε 0 pointing to the top. Therefore, the particle will experience the
Coulomb force FC = eE = eσ / 2ε 0 pointing to the top and the gravitational force F = mg
pointing to the bottom. To keep the particle above the ground, the total force should be
equal zero, i.e. mg = eσ / 2ε 0 from which we can find the value of
σ = 2ε 0mg / e = 2 × 8.85 × 10 −12 × 1 × 10 −6 × 9.89/1 × 10−3 = 1.75 × 10−13C / m2
Answer C.
Problem 10. Three particles are placed at the corners of the equilateral triangle with the
side a=1 cm. Top particle has a positive charge of 3 C and two bottom particles have a
negative charge of –3 C. What are the value and the direction of the field exactly at the
center of the triangle?
A) 0.17 x 10 -16 N/C, top
B) 2.35 x 10-6 N/C, top
C) 8.55 x 10-1 N/C, bottom
D) 1.6 x 10 15 N/C, bottom
E) Zero
Solution. Top particle has a positive charge, therefore it will create an electric field
pointing to the bottom. Each bottom particle will create an electric field which points
towards it, therefore the sum of the two fields will also point to the bottom. The total field
which is a sum of three fields will point to the bottom. Let us choose our x axis from the
left to the right and y-axis from the top to the bottom with the origin exactly at the center
of the triangle. We need to find a projection of each electric field onto y-axis. The value
2
of electric field from the first particle is E1 = k | q1 | / d and its projection onto y axis
E y1 = E1 = k | q1 | / d 2 where d is the distance between the center of the triangle and its
corner which is d = a / 3 . The value of electric field from each bottom particle is
E2,3 = k | q2,3 | / d 2 , and its projection onto y axis is E y 2,3 = E2,3 sin60 = E 2,3 / 2 . The total
electric field projected onto y axis is
E y tot = E y 1 + E y2 + E y3 = k | q1 | / d 2 + k | q2 | / 2d 2 + k | q3 | / 2d 2 =
8.99*109 *3/(10−2 / 3)2 *(1 + 0.5 + 0.5) = 1.6*1015 N / C
Answer D
Problem 11. A positively charged particle with q= 1mC and m=10 mg is hung on a wire
at some distance from a large positively charged plate with some surface charge density.
Find the surface density if the angle is known to be 45 degrees.
A) 1.76 x 10 -16 C/m2
B) 1.76 x 10-12 C/m2
C) 1.76 x 10-8 C/m2
D) 1.76 x 10 -4 C/m2
E) 1.76 C/m2
Solution. The plate creates an electric field pointing to the left, therefore the particle
experiences the Coulomb force FC = qE = qσ / 2ε 0 pointing to the left. The particle also
experiences the gravitational force Fg = mg pointing to the bottom and the reaction force
N pointing to the connection point. Since the particle is at the equilibrium, the sum of all
r
r r
three forces should be equal zero, i.e. FC + mg + N = 0 . Let us choose x axis from the left
to the right, and the y axis from the bottom to the top with the origin at the center of the
r
r r
F
+
mg
+ N = 0 on to y axis gives
particle. Projection of the equation C
N y = N cosθ = mg while projection on to x-axis gives qσ / 2ε 0 = N x = N sin θ from
which we can find σ = 2ε 0 N sin θ / q = 2ε 0 mg sin θ / q cosθ = 2ε 0mg tan θ / q which is
2*8.85*10 −12 *10*10−6 *9.95*1/10 −3 = 1.76*10−12 C / m2
Answer B.
3. GAUSS LAW (Chapter 24)
Key concepts:
Gauss Law is equivalent to the Coulomb law but sometimes more useful. Gauss law
considers a flux of an electric field thru some hypothetical surface (called Gaussian
surface) The flux of the field E thru the surface S is formally an integral
r r
Φ = ∫ EdS
The Gauss law relates the flux of the electric field thru the Gaussian surface with the total
charge enclosed by this surface
r r q
Φ = ∫ EdS = total
ε0
The usefulness of the Gauss law is seen when calculating electric field from some
symmetrical objects. For these situations, the electric field can for example be a constant
on the surface of the integration and can be taken out from the sign of the integral. The
integral can be taken easily (it is just the area of the Gaussian surface) and one can
immediately figure out the electric field of such object.
2
The units of electric flux: Nm / C
The Gauss law helps to understand the behavior of the electric field inside the
conductors. Since the conductors are the objects where the electrons travel freely, there is
no electric field exist inside the conductor. In other words, if a conducting object is
placed inside some external electric field, the charge inside the conductor will move
towards its surface until the electric field created by this displaced charge will completely
compensate the external electric field. The total electric field inside the conductor is
therefore zero.
The Gauss law also helps to understand the distribution of the electric charge placed into
the conductor. Since the electric field inside every conductor is always zero, there cannot
be extra electric charge located inside the conductor. Therefore, any extra charge placed
into the conductor will be distributed only on the surface of the conductor. This can be
proven by considering any Gaussian surface lying completely inside the conductor, since
electric field inside the conductor is zero, there is no charge which is enclosed inside this
Gaussian surface.
Typical problems related to Gauss Law:
Problem 12. Find a flux thru a spherical Gaussian surface of the radius a = 1 m
surrounding a charge of 8.85 pC.
A) 1 x 10 -16N m2/C
B) 1 x 10-12 N m2/C
C) 1 x 10-8 N m2/C
D) 1 x 10 -4 N m2/C
E) 1 N m2/C
Solution. A flux thru the Gaussian surface is the charge located inside the surface.
2
− 12
−12
Therefore: Φ = q / ε 0 = 8.85*10 /8.85*10 = 1Nm / C
Answer E
Problem 13. A positive charge Q= 8 mC is placed inside a neutral spherical conducting
shell with the inner radius a and the outer radius b. Find the charges induced at the inner
and outer surfaces of the shell.
A) Inner charge = –8 mC, Outer charge = +8 mC
B) Inner charge = +8 mC, Outer charge = -8 mC
C) Inner charge = 0 mC, Outer charge = +8 mC
D) Inner charge = –8 mC, Outer charge = 0 mC
E) Inner charge = 0 mC, Outer charge = 0 mC
Solution. Inside any conductor electric field should be equal zero. Therefore, the electric
field crerated by the external charge inside the shell is compensated by the electric field
created by the induced charged at the inner and outer surfaces of the shell. Choosing the
Gaussian surface to enclose the external charge and the inner surface of the shell and
Φ = ∫ EdS = 0 = Q + qin
since the electric field inside the
conductor is zero. This gives us qin = −Q = −8mC
Since the conducting shell is electroneutral, the charge induced at the inner part of the
shell is taken out from the outer surface of the shell which
gives qin + qout = 0 => qout = 8mC
Answer A.
applying the Gauss law:
Problem 14. A poistive charge Q=8 mC is placed inside a spherical conducting shell
with the inner radius a and the outer radius b which has an extra charge of 4 mC. Find the
charges at the inner and outer surfaces of the shell.
A) Inner charge = –8 mC, Outer charge = +8 mC
B) Inner charge = +8 mC, Outer charge = -8 mC
C) Inner charge = +8 mC, Outer charge = -12 mC
D) Inner charge = –8 mC, Outer charge = 12 mC
E) Inner charge = -4 mC, Outer charge = +4 mC
Solution. Inside any conductor electric field should be equal zero. Therefore, the electric
field created by the external charge inside the shell is compensated by the electric field
created by the induced charged at the inner and outer surfaces of the shell. Choosing the
Gaussian surface to enclose the external charge and the inner surface of the shell and
applying the Gauss law:
Φ = ∫ EdS = 0 = Q + qin
since the electric field inside the
conductor is zero. This gives us Q = − qin = −8mC
The conducting shell here is not electroneutral, but has some extra charge 4 mC.
Therefore only portion of the charge induced at the inner part of the shell should be taken
out from the outer surface of the shell. Electroneutrality condition is
qin + qout = 4 mC => qout = 12 mC
Answer D.
Problem 15. Find the value of the electric field at a distance r= 10 cm from the center of
the non conducting sphere of the radius R= 1 cm which has an extra positive charge equal
to 7 C uniformly distributed among the volume of the sphere.
A) 6.3 x 10 12 N/C
B) 7.5 x 10-6 N/C
C) 1.2 x 10 0 N/C
D) 9.1 x 10 -3 N/C
E) 8.5 x 10 2 N/C
7
Solution. If the charge at the sphere is distributed uniformly, for r>R the electric field
created by it is the same as the electric field of the same amount of charge located at the
center of the sphere.
q
7
E = k 2 = 8.99*10 9
= 6.3*1012 N / C
r
(10*10 −2 )2
Answer A.
Problem 16. A positive charge is placed inside a spherical metallic shell with inner
radius a and outer radius b. The charge is placed at shifted position relative to the center
of the shell. Describe the charge distribution induced at the shell surfaces.
A) A negative charge with uniform surface density will be
induced at the inner surface, a positive charge with the uniform
density will be induced at the outer surface.
B) A negative charge with non-uniform surface density will be
induced at the inner surface, a positive charge with the nonuniform density will be induced at the outer surface.
C) A positive charge with uniform surface density will be
induced at the inner surface, a negative charge with the uniform density will be induced
at the outer surface.
D) A positive charge with non-uniform surface density will be induced at the inner
surface, a negative charge with the non-uniform density will be induced at the outer
surface.
E) A negative charge with uniform surface density will be induced at the inner surface, a
positive charge with the non-uniform density will be induced at the outer surface.
Solution. First, since the external charge placed inside has a positive sign, the inner
surface will be charged negatively, and the outer surface will be charged positively.
There is no charge induced in the bulk of the metallic shell. Second since the external
charge is not precisely at the center of the shell, the induced charge densities at the inner
and outer surfaces will not be uniform.
Answer B.