Electrostatics: Electric potential

1. Introduction

We have defined the vector electric field E as the vector electric Coulomb's force F per unit of test charge q_o: E = F/q_o.

We introduce another field called electric potential or simply potential V defined as a electric potential energy per unit charge.

The potential energy is a scalar, so is the potential field V. The vector electric field E and the scalar electric potential V are derivable from to another. The electric potential difference between two points is known as the voltage measured in Volt. The SI unit of the electric potential energy is Joule, therefore the SI unit of the potential is Joule/Coulomb = Volt.

2. Work done by electric force on
test charged particle

Consider a test charged particle q_o moving in the electric potential field due to the point charge q. The electric force on the test charge is F = E/q_o. The two vectors F and E are parallels.

2.1. Along circular arc

1. The work done by the electric force F on the test charged particle moving along a circular arc, from a to b, centred at the fixed point charge q is:
W = ∫ F. dl = ∫ F dl cos (F, dl) = 0, because dl is perpendicular to E at each point on the path.

2.2. Along radial path

2. The work done by the electric force F on the test charged particle moving along a radial path , from a to b is:
W = ∫ F. dl = ∫ F dl , because dl is parallel to E at each point on this radial path.

F = (1/4πε_o) qq_o/r²

E = F/q_o

The infinitesimal displacement dl = dr ^r, therefore

W = q_o∫ E. dl = (1/4πε_o) qq_o/r² dr

	rb
W = (qqo/4πεo) ∫		1/r2 dr   = (qqo/4πεo) (1/ra - 1/rb)
	ra

2.3. Along radial-arc path

The path from a to b encounters arc-path with W = 0 and radial-path with W = ∫F.dl. Since the the terms between those related to r_a and r_b cancel out, the contributions of all the arc-paths is zero and of all the radial-paths gives:

W = (qq_o/4πε_o) (1/r_a - 1/r_b)

The result is independent on any point between a and b.

2.4. Along arbitrarily shaped path

Any arbitrarily shaped path, between a and b, can be viewed as an infinite number of infinitesimal arc paths and radial paths. Only the radial-paths contributions survive and the work done on the particle test by the force F between a and b is again dependent only on 1/r_b and 1/r_b.

The work done on a small test charged particle by an electric force in an electric field due to a fixed point charge depends only on the separation distance between the particles before the displacement (ra) and and after the displacement (rb).

The work done is independent of the path. If the work done by a force is independent of the path, the force is conservative, and the change in potential energy is the negative of the work done by the conservative force :

	b
W = ∫		F.dl = Ua - Ub = - (Ub - Ua) = - ΔU
	a

3. Electric potential energy

3.1. Test particle in the field of a point charge

The work done by an electric force, which is conservative, in an electric field, due to a point fixed charge q, on a test charged particle q_o moving from a to b is equal to the negative of the change in the test particle's potential energy :

W = - ( U_b - U_a) = U_a - U_b, with

U_a = (qq_o/4πε_o) (1/r_a)

And

U_a = (qq_o/4πε_o) (1/r_b)

We can find a general expression of U at any point selecting a reference position at which the potential energy is zero. The simple choice is r_a = ∞ and r_b = r, so

U_a = 0 and U_b = U(r), that is

	r
U(r) = - ∫		F.dl = qoq/4πεor
	∞

U(r) is the electric potential energy of the charged test particle q_o on the electric field due to the fixed charge point + q.

3.2. Test particle in the field of a number of point fixed charges

Consider a test particle qo moving from a to b in the field due to two points fixed charges q1 and q2 distant r1 and r2 from qo respectively.

According to the principle of superposition, the work done on the test particle by the net electric force conservative F of the two contributions of the field due to q1 and q2 is:

	b			b			b			b
W = ∫		F.dl =	∫		qo (E1 + E2).dl =	qo∫		E1.dl +	qo∫		E2.dl
	a			a			a			a

The corresponding potential energy is U = - W:

U = U1 + U2 = (qo/4πε_o) (q1/r1 + q2/r2)

r1 and r2 are the distances between the test charge and the two fixed point charge q1 and q2 respectively.

Extending to a case of more that two fixed particles, we have:
U = (qo/4πε_o) Σqi/ri
Where ri is the distances between the test charge and the fixed point charge i.

3.3. Potential energy in a constant field

The negative of the work done on a moving charge q from a to b, by an electric force F, due an electric field E is the change in the elctric potential energy of the this charge:

	rb
- W = Ub - Ua = - ∫		F.dl = q E Δl
	ra

If E is constant in the z direction

U_b - U_a = - q E (z_b - z_a)

At a point z, we write:

U_z = - q E z

For a dipole of dipole moment: p = d q ,
the potential energy is U = U(+ q) + U(- q) =
- q E z + - (- q) E (- z) = - 2 q E z = - 2 q E (d/2) cos θ = - q d E cos θ =
- p E cos θ = - p . E

		→ →
U = -		P . E

4. Electric potential

4.1. Definition

We defined the electric field E of a fixed charge q as the ratio of the electric force exerted on a small charged particle test and its charge qo, by E = F/qo. It is important that the charge qo used to measure the electric field E to be small to not alter the charge distribution that produced the potential due to the fixed charge q.

Similarly, we define the electric potential or potential V as the ratio of the electric potential energy U and the charge test qo.

V = U/qo

From the above expression of the electric potential energy, we have:

V(r) = U(r)/qo = q/4πε_or
for a single fixed charge, and

V(r) = U(r)/qo = (1/4πε_o) Σqi/ri
for a distribution of fixed charged particles

V(r) as U(r) is a scalar quantity. The potential V(r) has a value at each point in space where the fixed charge is present. The SI unit of U(r) is the Joule. The SI unit of the potential is the Joule/Coulomb = Volt (J/C = V).

Once we know the value of the potential V(r) at the point at the distance r from the fixed charge producing V(r), we can determine the potential energy U(r) = qo V(r) for a charge qo.

The charge of elementary particles as electrons and protons is small. The charge of electron is e = 1.6 x 10^{- 19} Coulomb. From this value, we define the unit of energy called the electron-Volt (eV).
1 eV = 1.6 x 10^{- 19} J.

4.2. Example 1: Distribution of three charges

1 N.m/C = 1 J/C = 1 V
1 nC = 1 x 10^-9 C

q1 = + 5 nC, q2 + 3 nC, and q3 = - 4 nC.
a = 3 cm , b = 4 cm → r = 5 cm

V(r) = k (3/3 - 4/4 + 5/5 ) =
9.0 x 10⁹N.m²/C² . 1 nC/cm =
9.0 x 10⁹N.m²/C² . 1 x 10^-9C/10^-2m =
9.0 x 10² V = 900 V.

4.3. Example 2: Dipole's potential

The dipole of moment p = 2dq k^{^} located at the origin. We want to determine the potential produced by the dipole far from the two charges + q and - q , that is at a distance r greater than the separation 2d between the two charges.

At the point P, we have:

V(r) = (1/4πε_o) Σqi/ri =
V(r) = (1/4πε_o) q(1/r1 - 1/r2)

Far from the dipole: r ≫d, we can write:

r1 ≈ r2 ≈ r, and
r2 ≈ r1 + 2 d cos θ

Therefore

V(r) = (1/4πε_o) 2 q d cos θ(r2 - r1)/r1r2 ≈
(1/4πε_o) 2q d cos θ/r² = (1/4πε_o) p cos θ/r²
= (1/4πε_o) p . r^{^} /r²

At θ = 90° V(r) = 0. The potential is zero at all points in the plane xy, positive at all points above xy plane, and negative at all points below the xy plane.

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Potential due to a dipole:
V(r) = (1/4πε_o) 2q d cos θ/r²

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4.4. Potential due to continuous charge distribution

The potential for a number of point fixed charges is:

V(r) = (1/4πε_o) Σqi/ri

We transform this equation into the equation for a continuous charge distribution. The continuous charge distribution is considered as a large number "n" of individual fixed charges qi. In the limit when n tends to infinity, the sum becomes an integral:

			n
V = lim	(1/4πεo)		Σ	qi/ri	= (1/4πεo)∫		dq/r
n → ∞			i=1

r is the distance from dq to the point at which the potential is evaluated.

4.5. Potential due to uniformly charged circular ring

Consider a uniformly charged circular ring of radius R and total charge Q.

4.6. Potential due to a thin uniformly charged disk

Consider a uniformly charged disk of radius R and surface cahrge density σ. .

5. Potential difference

5.1. Definition

We have defined the potential energy U with the reference position at ∞, where U = 0. The potential V is also zero at ∞, and takes also r = ∞ as the reference position.

Note that a potential energy or potential are just mathematical functions, but a change in potential energy and a change in potential has a physical significance. (an abject in a free-fall motion has a potential energy related to a certain reference position).

The potential difference between two points, that is the change in potential does not need to use the reference position V_∞ = 0.

If Ub - Ua represents the electric potential energy difference of a small test charge qo moving from a point a to a point b in the field due a fixed charge distribution, then the corresponding potential difference is:

Vb - Va = (Ub - Ua)/qo

The difference in the electric potential energy of a test charge qo is the negative of the work done by the electric force on this charge.

So

	b
Ub - Ua = - qo ∫		E.dl
	a

Therefore

	b
Vb - Va = - ∫		E.dl
	a

Further, we see that the SI unit of the potential is also the Volt/meter (V/m).

5.2. Application: Uniform electric field

The potential difference between two points a and b in a uniform field E along the x-axis is:

	b
Vb - Va = - E ∫		dl = - E (xb - xa)
	a

5.3. Potential difference in a parallel-plate capacitor

Conside a capacitor formed by two parallel plates. The potential difference between the two plates is the same as between two points a and b in a uniform field E along the x-axis :

	b
Vb - Va = - E ∫		dl = - E (xb - xa) = - E d
	a

The lateral dimension of each plate are much larger than the plate separation d between the plates. the electric fiels E is then uniform and it is equal to σ/ε_o. Therefore V(x) = V_a - V_b = E d

5.4. Potential of a uniformly charged sphere

Consider a sphere uniformly charged of radius R and charge Q.

We have determined the expression of the elctric field due to a uniformly charged sphere:

E(inside) = Qr/4πrε_oR³ = (ρ/3ε_o) r

E(outside) = Q/4πε_o(1/r²) = ρR³/3ε_o(1/r²)

Therefore, the expression of the related potential are:

V(inside)

	b
Vb - Va = - ∫		Qr/4πεoR3 dr = Q/8πεoR3[ra2 - rb2]
	a

V(outside)

	b
Vb - Va = - ∫		Q/4πεo(1/r2) dr = Q/4πεo[1/rb - 1/ra]
	a

6. Electric field and potential

6.1. Definition

The potential is defined as

	r
V(r) = - ∫		E.dl
	∞

r = ∞ is the reference position that corresponds to V_∞ = 0

The position r is the distance between the fixed charge and the test charge, that is where the potential is avaluated.

Thus the potential is the negative of the line integral of the electric field from ∞ to where is evaluated.

We are interested in the reverse operation. Given a potential V, how tho determine the corresponding electric field E?

Consider an infinitesimal change of the potential in the x direction dV = V(x + dx) - V(x). The electric field E is nearly constant and the above integral becomes:

	x+dx
dV = - Ex ∫		dx = - Ex(x' + dx' - x') = - Ex dx
	x

Thus

dV = - E_x dx
with y and z held constant. Therefore

∂V - E_x ∂x

Similarly

∂V = - E_y ∂y, and
∂V = - E_z ∂z

So, the components of E are given by the negative of the partial derivatives of V.

In the case of a charge distribution with spherical symmetry, V depends only on the radial coordinate r, so E has only a radial coordinate, we have:

∂V = - E_r ∂r

6.2. Example Electric dipole

We have found the expression of the potential for a dipole of moment p = 2dq k^{^}, V(r) = (1/4πε_o) 2q d cos θ/r²

In cartesian ccordinates

r² = x² + y² + z² cos θ = z/r = z/[x² + y² + z² ]^1/2

The expression of V becomes:

V(r) = (1/4πε_o) p z/[x² + y² + z² ]^3/2

We have
- ∂V/∂x = E_x = (p/4πε_o) 3xz/r⁵
- ∂V/∂y = E_y = (p/4πε_o) 3yz/r⁵
- ∂V/∂z = E_z = (p/4πε_o) (3z²/r⁵ - 1/r³)

6.3. Other expression of the electric
field around a dipole:

E = - ∇V

That is

E = - (∂V/∂x, ∂V/∂y, ∂V/∂z)

V = (1/4πε_o) p.r/r³

= (1/4πε_o) (xp_x + yp_y + zp_z)/ (x² + y² + z²)^3/2

Differentiating with respect to x, gives:

∂V/∂x = (1/4πε_o) [p_x(x² + y² + z²)^3/2 - (3/2)(2x) (xp_x + yp_y + zp_z) (x² + y² + z²)^1/2]/ (x₂ + y₂ + z₂)³ =
(1/4πε_o)[ p_x/(x² + y² + z^23/2 - (3/2)(2x) (xp_x + yp_y + zp_z)/ (x² + y² + z²)^5/2]

∂V/∂x = (1/4πε_o)[p_x/r³ - (3x)(p.r)/r⁵]

Similarly,

∂V/∂y = (1/4πε_o)[p_y/r³ - (3x)(p.r)/r⁵]

∂V/∂z = (1/4πε_o)[p_z/r³ - (3x)(p.r)/r⁵]

Therefore

E = - ∇V = - (1/4πε_o)[p/r³ - (3r)(p.r)]/r⁵

E = - ∇V = - (1/4πε_o)[p - 3(p.r^)r^]/r³

r^ is the unit vector position.

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Electric field around a dipole:

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