Electrostatic force, field, potential energy, potential
Point charges
Problem:
Consider 3 positive charges q at the vertices of an equilateral triangle.
Each side has length l. Find the total force (magnitude and direction) on each of the charges.
Solution:
 Concepts:
Coulomb's law, F_{12} = (k_{e}q_{1}q_{2}/r_{12}^{2})
(r_{12}/r_{12}),
k_{e} = 1/(4πε_{0}) = 9*10^{9} Nm^{2}/C^{2},
the principle of superposition
 Reasoning:
We are asked to calculate the force exerted on a point charge by other
charges.
 Details of the calculation:
Arrange the charges as shown in the figure.
The force on q_{1} is
F_{1 }= F_{21 }+ F_{31} (vector
sum).
The horizontal components of F_{21 }and F_{31}
cancel and the vertical components add. F_{1} points upward, in
the ydirection. The magnitude of F_{1} is F_{1 }= F_{21}cos30^{o
}+ F_{31}cos30^{o}.
The magnitudes F_{21} and F_{31} are given by F_{21 }= F_{31
}= k_{e}q^{2}/l^{2}.
F_{1 }= 2 k_{e}(q^{2}/l^{2})cos30^{o} in
the ydirection, F_{1 }= 2 k_{e}(q^{2}/l^{2})cos30^{o}
j.
F_{2} and F_{3} have the same magnitude. If F_{1}
points north, then F_{2} points 30^{o} south of west and
F_{3} points 30^{o} south of east.
Problem:
A small negatively charged ball of mass m is suspended on a long insulating
string. An external force very slowly moves another small negatively
charged ball on a horizontal path towards the first one from a large distance.
Eventually, the second ball reaches the original location of the first one.
At that moment, the first ball is elevated a small distance h above its original
position. How much work is done by the external force moving the second
ball to the original location of the first one?
Solution:

Concepts:
Work and energy

Reasoning:
The ball moves very slowly, we can neglect radiation. The work done to
the system is equal to the sum of the increase of the gravitational
potential energy of the first ball and to the electrostatic potential energy
stored in the pair of charged balls.

Details of the calculation:
W = ∆U_{g} + ∆U_{e} = mgh + k_{e}q_{1}q_{2}/d,
where k_{e} = 1/(4πε_{0}) and d is the final distance
between the balls.
In equilibrium, the resultant of the weight mg and
the Coulomb force along the string is canceled the tension in the string. The components perpendicular to the string must cancel each other.
From geometry:
(k_{e}q_{1}q_{2}/d^{2})cos(θ/2)
= mgcos(90^{o}  θ) = mg sinθ = 2mg sin(θ/2)cos(θ/2)
(k_{e}q_{1}q_{2}/d^{2}) = 2mg sin(θ/2)
We also have
sin(θ/2) = h/d.
Therefore (k_{e}q_{1}q_{2}/d^{2}) = 2mg h/d, U_{e} = k_{e}q_{1}q_{2}/d = 2 mgh.
W = 3 mgh is the work done by the external agent.
Problem:
A proton is released from rest at a distance of 1 Angstrom from another proton.
(a) What is the total kinetic energy when the protons have moved infinitely far
apart?
(b) What is the terminal velocity of the moving proton if the other is kept at
rest? If both are free to move, what is their velocity?
Solution:
 Concepts:
Energy conservation
 Reasoning:
The Coulomb force is a conservative force.
 Details of the calculation:
(a) E = T_{final} = kq^{2}/r_{initial} =
9*10^{9}*(1.6*10^{19})^{2}/1*10^{10} J = 2.3*10^{18} J = 14.4 eV.
(b) If one proton is fixed, the terminal speed of the other proton is
v = (2E/m)^{1/2} = 5.25*10^{4} m/s. The free proton moves away from the
fixed proton.
If both protons are free to move, they share the kinetic energy equally.
The speed of each proton is v = (E/m)^{1/2} = 3.71*10^{4} m/s. The
protons move away from each other.
Problem:
An alpha particle containing two protons is
shot directly towards a platinum nucleus containing 78 protons from a very
large distance with a kinetic energy of 1.7*10^{12 }J.
The platinum nucleus is held fixed in a matrix.
What will be the distance of closest approach?
Solution:
 Concepts:
Energy conservation
 Reasoning:
The electrostatic force is a conservative force. (We are
neglecting radiation.)
 Details of the calculation:
The electric charge of the alpha particle is q_{1 }= 2q_{e} and
that of the platinum nucleus is q_{2 }= 78q_{e}. The alpha particle
and the nucleus repel each other. As the alpha particle moves towards the
nucleus, some of its kinetic energy will be converted into electrostatic
potential energy. At the distance of closest approach, the alpha
particle's velocity is zero, and all its initial kinetic energy has been
converted into electrostatic potential energy. The initial kinetic energy
of the alpha particle is KE = 1.7*10^{12 }J.
The final potential energy is U = kq_{1}q_{2}/d, where d
is the distance of closest approach.
We have k_{e}q_{1}q_{2}/d
= 1.7*10^{12 }J.
This yields
d = k_{e}q_{1}q_{2}/(1.7*10^{12 }J)
= (9*10^{9}*4*78*(1.6*10^{19})^{2}/(1.7*10^{12})) m
= 2.1*10^{14 }m
for the distance of closest approach.
The alpha particle's velocity
is zero at d. The acceleration is in a direction away from the
nucleus. The distance between the alpha particle and the nucleus
will increase again, converting the potential energy back into kinetic
energy.
Problem:
A proton is fired directly at alpha particle (He^{2+}) such that the two
particles are initially approaching one another with the same (nonrelativistic)
speed v_{0} when they are far apart. What is the classical distance of
closest approach of the two particles?
Solution:
 Concepts:
Conservation laws
 Reasoning:
Transforming to the CM frame simplifies the problem.
 Details of the calculation:
Energy conservation
In their CM frame the particles approach each other with momenta that have equal
magnitudes.
v_{He} = 2v_{0}/5 and v_{p} = 8v_{0}/5
in the CM frame.
At the distance of closest approach d the kinetic energy of both particles has
been completely converted into electrostatic potential energy.
½ 3m (2v_{0}/5)^{2} + ½ m (8v_{0}/5)^{2} =
(8/5)mv_{0}^{2} = k_{e}2q_{e}^{2}/d,
d = 5k_{e}q_{e}^{2}/(4mv_{0}^{2}) = 5q_{e}^{2}/(16mπε_{0}v_{0}^{2}).
(k_{e} = 1/(4πε_{0})).
Or:
Momentum conservation: 4mv_{0}  mv_{0} = 5mv', v' = (2/5)
v_{0}
(At the distance of closest approach the two particles instantaneously move with
the same speed v', the speed of the CM. They have no relative velocity.)
Change in total energy of the particles: ΔE = ½ 5m(v_{0}^{2}
 v'^{2}) = (8/5)mv_{0}^{2} = k_{e}2q_{e}^{2}/d.
Problem:
Two point charges +q and q are connected by a spring with spring constant k
and equilibrium length L.
(a) The system is in equilibrium if the distance between the charges is
L/2. In equilibrium, what is the electric dipole moment p of the
system in terms of k and L?
(b) At large distances away from the system (r >> L) what is the electric field
E(r) produced by the system (1st order)?
For the case of p being located at the origin and pointing in the
zdirection, write E(r) in terms of spherical coordinates and unit
vectors.
Solution:
 Concepts:
The electric dipole, equilibrium
 Reasoning:
The dipole moment of a positive point charge q
displaced by d with respect to a negative point charge q is
p = qd.
 Details of the calculation:
(a) In equilibrium q^{2}/(πε_{0}L^{2}) = kL/2.
q = (kπε_{0}L^{3}/2)^{½}.
Choose your coordinate system so that the charges lie on the zaxis and place the
origin midway between the charges.
Then p = qL/2 k = (kπε_{0}L^{5}/8)^{½}
k.
(b) With the midway point located at the origin, the electric field at
large distances is E(r) = [1/(4πε_{0}r^{3})][3(p·r)r/r^{2}
 p].
With p = p k we have
E(r) = [p/(4πε_{0}r^{3})][2cosθ e_{r}
+ sinθ e_{θ}]
= [(kπε_{0}L^{5}/8)^{½}/(4πε_{0}r^{3})][2cosθ
e_{r} + sinθ e_{θ}]
= (kL^{5}/(128πε_{0}r^{6}))^{½}[2cosθ
e_{r} + sinθ e_{θ}].
Continuous charge distributions
Problem:
A vertical thin rod of length L carries a total charge Q uniformly
distributed. Calculate the electric field along its axis at a distance z above
its top end.
Solution:
 Concepts:
The electric field due to a charge distribution, the
principle of superposition
 Reasoning:
We are asked to find the electric field due to a line
charge distribution.
 Details of the calculation:
E(z) = k∫_{L}^{0} λdz'/(z 
z')^{2}
= kλ∫_{z+L}^{z} dz''/z''^{2} = kλ(1/z  1/(z + L)) = kQ/((z + L)z).
E = Ek.
Problem:
Consider a line charge with line charge density λ
= Q/2a that extends along the xaxis from x = a to x = +a. Find the electric
field on the yaxis.
Solution:
 Concepts:
The electric field due to a charge distribution, the principle of
superposition
 Reasoning:
We are asked to find the electric field due to a line charge distribution.
 Details of the calculation:
The field on the yaxis due to an infinitesimal element of charge
λdx is given by
dE_{x} = (k_{e}λdx/(x^{2} + y^{2}))
cosθ, dE_{y} = (k_{e}λdx/(x^{2} + y^{2}))
sinθ,
where cosθ = x/(x^{2} + y^{2})^{½}, sinθ = y/(x^{2}
+ y^{2})^{½}.
On the yaxis the field due to the line charge therefore is given by
E_{x} = k_{e}λ∫_{a}^{a}xdx/(x^{2}
+ y^{2})^{3/2} = 0 from symmetry, and
E_{y} = k_{e}λy∫_{a}^{a}dx/(x^{2}
+ y^{2})^{3/2}.
Using ∫_{a}^{a}dx/(x^{2} + y^{2})^{3/2}
= 2a/(y^{2}(a^{2} + y^{2})^{½})
we have E_{y} = k_{e}λ 2a/(y(a^{2} + y^{2})^{½})
= k_{e}Q/(y(a^{2} + y^{2})^{½})
The electric field on the yaxis points in the positive ydirection for y > 0
and in the negative ydirection for y < 0.
As y becomes very large, we can neglect a^{2 }compared to y^{2}
under the square root and then E_{y }= k_{e}Q/y^{2 }
∝ 1/y^{2}. From very far away, the line
charge looks like a point charge.
If, on the other hand, the line is very long, and we y << a, then we can neglect
y^{2 }compared to a^{2} under the square root and then E_{y
}= k_{e}Q/(ay) ∝ 1/y. Near a very long
line charge the electric field falls off as 1/distance, not as 1/distance^{2}.
Problem:
A disk of radius a carries a nonuniform surface charge density given by
σ = σ_{0} r^{2}/a^{2}, where σ_{0} is a
constant.
(a) Find the electrostatic potential at an arbitrary point on the disk
axis, a distance z from the disk center and express the result in terms of
the total charge Q.
(b) Calculate the electric field on the disk axis and express the result in
terms of the total charge Q.
(c) Show that the field reduces to an expected form for z >> a.
(d) To first order in ρ, find an expression for the radial component of
E(ρ, φ, z) at a distance ρ << a away from the zaxis and evaluate it for
z >> a.
Solution:
 Concepts:
The electrostatic potential
 Reasoning:
We find the electrostatic potential Φ(r) due to a charge distribution using
the principle of superposition.
 Details of the calculation:
(a) Φ(r) = k ∫_{A'}σ(r)dA'/r 
r'
We are asked to find the potential due to a surface charge distribution
in a situation with symmetry.
dΦ(z) = k2πrσ(r)dr/(r^{2 }+ z^{2})^{½} = (k2πσ_{0}/a^{2})r^{3}dr/(r^{2
}+ z^{2})^{½} is the potential on the axis due
to a ring of radius r.
Here k = 1/(4πε_{0}).
Φ(z) = (k2πσ_{0}/a^{2})∫_{0}^{a} r^{3}dr/(r^{2
}+ z^{2})^{½}.
Q = (2πσ_{0}/a^{2})∫_{0}^{a} r^{3}dr
= ½(kπσ_{0}a^{2}), therefore Φ(z) = (4Q/a^{4})∫_{0}^{a}
r^{3}dr/(r^{2 }+ z^{2})^{½}.
Φ = (4kQ/a^{4})[(1/3)(a^{2 }+ z^{2})^{3/2}
 z^{2}(a^{2 }+ z^{2})^{½} + 2z^{3}/3].
(b) On the axis, the electric field only has a zcomponent,
E =
E_{z} k.
E_{z}(z) = ∂Φ/∂z = (4kQ/a^{4})[z(a^{2 }+ z^{2})^{½}
 z^{3}(a^{2 }+ z^{2})^{½} + 2z^{2}]
= (4kQ/a^{4})[(z^{2}(1 + a^{2}/z^{2})^{½}
+ z^{2}(1 + a^{2}/z^{2})^{½}  2z^{2}]
= (4kQz^{2}/a^{4})[(1 + a^{2}/z^{2})^{½}
+ (1 + a^{2}/z^{2})^{½}  2]
(c) If z >> a then to 2nd order in a^{2}/z^{2} we have
(1 + a^{2}/z^{2})^{½} = [1 + ½a^{2}/z^{2}
 (1/8)(a^{4}/z^{4})]
(1 + a^{2}/z^{2})^{½} = [1  ½a^{2}/z^{2}
+ (3/8)(a^{4}/z^{4})]
Therefore to second order [(1 + a^{2}/z^{2})^{½}
+ (1 + a^{2}/z^{2})^{½}  2] = ¼(a^{4}/z^{4}),
E_{z}(z) = kQ/z^{2} = field of a point charge Q.
(d) ∇∙E = 0 for z ≠ 0,
E has no
φ dependence from
symmetry.
In cylindrical coordinates: (1/ρ)∂(ρE_{ρ})/∂ρ + (1/ρ) ∂E_{φ}/∂φ
= ∂E_{z}/∂z.
E_{φ} = 0, so ∂(ρE_{ρ})/∂ρ = ρ∂E_{z}/∂z.
E_{ρ} + ρ∂E_{ρ}/∂ρ = ρ∂E_{z}/∂z.
Near the axis (ρ << a) we expand this expression, keeping only first
order tems of the Taylor series.
E_{ρ}_{ρ=0} + ∂(E_{ρ})/∂ρ_{ρ=0}ρ
+ (ρ∂E_{ρ}/∂ρ)_{ρ=0} + ∂(ρ∂E_{ρ}/∂ρ)∂ρ_{ρ=0}ρ
= (ρ∂E_{z}/∂z)_{ρ=0}  ∂(ρ∂E_{z}/∂z)∂ρ_{ρ=0}ρ.
Several tems evaluate to zero,
∂(E_{ρ})/∂ρ_{ρ=0}ρ + ∂E_{ρ}/∂ρ_{ρ=0}ρ
= ∂E_{z}/∂z_{ρ=0}ρ.
∂(E_{ρ})/∂ρ_{ρ=0} = ½∂E_{z}/∂z_{ρ=0}.
E_{ρ} = ∂(E_{ρ})/∂ρ_{ρ=0}ρ
= ½∂E_{z}/∂z_{ρ=0}ρ = ½ ρ f(z),
where f(z) = ∂/∂z[(4kQz^{2}/a^{4})[(1 + a^{2}/z^{2})^{½}
+ (1 + a^{2}/z^{2})^{½}  2]].
For z >> a, E_{ρ} = (ρ/z) kQ/z^{2}.
Problem:
Charge is distributed
uniformly over a thin circular disk of radius R. The charge per unit area is
σ. Calculate the electric field and
potential due to the disk at a point P on the axis of the disk at a distance z
away from the center.
Solution:
 Concepts:
Electrostatics
 Reasoning:
We are asked to calculate the field and potential for a static charge
distribution.

Details of the calculation:
Let the disk lie in the xyplane with its center at the origin. Then the
zaxis is the symmetry axis.
dV(z) = k2πrσdr/(r^{2 }+ z^{2})^{½}
is the potential on the axis due to a ring of radius r.
Integrating from r = 0 to r = R we find
V(z) = k2πσ{(R^{2}+z^{2})^{½
} z} for a disk of radius R. Q =
σπR^{2}, so
V(z) = k(2Q/R^{2})[(R^{2}+z^{2})^{½
} z].
E_{z} = dV(z)/dz = k2πσ[1
 z/(R^{2}+z^{2})^{½}]. Symmetry dictates
that E = E_{z}k.
Similar problems
Unknown charge distributions
Problem:
The electric field in some volume V of space is E = [2A/(x^{3}y^{2})
 By]i + [2A/(x^{2}y^{3})  Bx]j.
(a) Show that the line integral of this field over a close path is zero.
(b) Which charge density ρ in the volume V is consistent with this field?
Solution:
 Concepts:
Conservative fields
 Reasoning:
Electrostatic fields are irrotational. If
∇×E
= 0, the line
integral of E over any close path is zero.
 Details of the calculation:
(a) (∇×E)_{x} =
∂E_{z}/∂y  ∂E_{y}/∂z
= 0, (∇×E)_{y} =
∂E_{x}/∂z  ∂E_{z}/∂x
= 0.
E has no zcomponent and the x and ycomponents do not depend on z.
(∇×E)_{z}
= ∂E_{y}/∂x  ∂E_{x}/∂y
= 4A/(x^{3}y^{3})  B  (4A/(x^{3}y^{3})  B)
= 0.
∇×E =
0, the line integral of
E over any close path is
zero. (Stokes' theorem)
(b) ρ/ε_{0} = ∇·E = ∂E_{x}/∂x + ∂E_{y}/∂y
= 6A/(x^{4}y^{2})  6A/(x^{2}y^{4}) = (6A/(x^{2}y^{2}))(1/x^{2}
+ 1/ y^{2})
ρ = 6Aε_{0}r^{2}/(x^{4}y^{4}), with r^{2}
= x^{2} + y^{2}.