Notes to a video lecture on http://www.unizor.com
Electric Field Potential
The general form of the Coulomb's Law, when two electrically charged point-objects, A and B, are involved, is
F = k·qA·qB / R²
F is the magnitude of the force of attraction (in case of opposite charges) or repelling (in case of the same type of charge, positive or negative) in newtons(N)
qA is electric charge of point-object A in coulombs(C)
qB is electric charge of point-object B in coulombs(C)
R is the distance between charged objects in meters(m)
k is a coefficient of proportionality, the Coulomb's constant, equals to 9.0·109 in N·m²/C²
We have introduced a concept of electric field intensity as a force acting on a probe point-object B, charged with +1C of electricity, from a field produced by the main object A. This force is a characteristic of a field at a point where a probe object is located and is equal to
E = k·qA / R²
If we want to move a probe point-object from one point in the field to another in uniform (without acceleration) motion, we have to take into account this force. It can help us to do the move, if this force acts in the direction of a motion, or prevent this motion, if it acts against it. In a way, the electric field becomes our partner in motion, helping or preventing us to do the move.
Work of Coulomb's Force
Of obvious interest is the amount of work needed to accomplish the move. If we act against the force of electric field intensity, we have to spend certain amount of energy to do the work. If the field force helps us, we do not spend any energy because the field does it for us. Similar considerations were presented in the Gravitation part of this course.
Recall from the Mechanics part of this course that the work of the force F, acting at an angle φ to a trajectory on the distance S, is
W = F·S·cos(φ)
For a non-uniform motion and variable force all components of this formula are dependent on some parameter x, like time or distance:
dW(x) = F(x)·dS(x)·cos(φ(x))
where we have to use infinitesimal increments of work dW(x) done by force F(x) on infinitesimal distance dS(x).
The angle φ(x) is the angle between a vector of force F(x) and a tangential to a trajectory at point S(x).
Using a concept of scalar product of vectors and considering force and interval of trajectory as vectors, the same definition can be written as
dW(x) = (F(x)·dS(x))
The latter represents the most rigorous definition of work.
Integration by parameter x from x=xstart to x=xend can be used to calculate the total work
W[xstart , xend ] performed by a variable force F(x), acting on an object in a non-uniform motion, on certain distance S(x) along its trajectory, as the parameter x changes from xstart to xend.
In case of a motion of an electrically charged object in an electrical field the force is the Coulomb's force.
Let's analyze the work needed to move such an object in the field of electrically charged point-object from one position to another.
Case 1. Radial Motion
Let the charge of the main point-object in the center of the electrical field be Q. We move a probe object of charge q along a radius from it to the center of a field from distance r1 to r2.
The Coulomb's force on a distance x from the center equals to
F(x) = k·Q·q / x²
The direction of this force is along the radial trajectory and the sign of the Coulomb's force properly describes whether the resulting work will be positive (in case of similarly charged main and probe objects, + and + or − and −) or negative (in case of opposite charges, + and − or − and +).
Since the force is variable and depends on the distance between the main object and the probe object, to calculate the work needed to move the probe object, we have to integrate the product of this force by an infinitesimal increment of the distance dx on a segment from r1 to r2.
W[r1,r2] = ∫[r1,r2]k·Q·q·dx / x²
Since the indefinite integral (anti-derivative) of 1/x² is −1/x, the amount of work is
W[r1,r2] = k·Q·q·(1/r1−1/r2)
This work is additive. If we move from a distance of r1 to a distance r2 and then from a distance r2 to a distance r3, the total work will be equal to a sum of works, which, in turn, would be the same as if we move directly to distance r3 without stopping at r2
W[r1,r2] + W[r2,r3] = W[r1,r3]
Case 2. Circular Motion
Consider now that we move a probe object circularly, not changing the distance from the center of the electric field.
In this case the vector of force (radial) is always perpendicular to the vector of trajectory (tangential). As a result, this motion can be performed without any work done by us or the field. So, for a circular motion the work performed is always zero.
Obviously, this work, as we move the probe object circularly, is also additive.
Case 3. General
Any vector of force in a central electric field can be represented as a sum of two vectors - radial, that changes the distance to a center of a field, and tangential (along a circle), which is perpendicular to a radius. That means that any infinitesimal increment of work can be represented as a sum of two increments - radial and tangential. Since the latter is always zero, the amount of work performed to facilitate this motion depends only on the distances to the center at the beginning and at the end of the motion.
The immediate consequence from this consideration is that the work needed to move a charged point-object from one point in the radial electric field to another is independent of the trajectory and only depends on starting and ending position in the field. Even more, for radial electric field it depends only on starting and ending distances to a center of the field.
This independence of work from trajectory is a characteristic not only of radial electric field, but of the whole class of the fields - those produced by conservative forces, and electrostatic forces are conservative. Gravitational forces are also of the same type.
As an example of non-conservative forces, consider an object moving inside the water from one point to another. Since the water always resists the movement, the longer the trajectory that connects two points - the more work is needed to travel along this trajectory.
Electric Field Potential
In the radial field produced by the point-object charged with Q amount of electricity the
Using the formula above for r1=∞ and r2=r for a probe object charged with q=+1C of electricity, we obtain the formula for an electric potential at distance r from a center of the field, where a point-object charged with Q amount of electricity is located
V(r) = −k·Q / r
Notice that the derivative of potential V(r) by distance r from a center gives the field intensity:
V'(r) = k·Q / r²
So, knowing the potential at each point of the radial field, we can determine the intensity at each point.
Since, as we stated above, the work performed to move a probe object in the electric field does not depend on trajectory, we can accomplish moving a probe object charged with +1C of electricity from distance r1 to distance r2 by, first, moving it to infinity, which results in amount of work
W = W1 + W2 = −V(r1)+V(r2) =
= k·Q·q·(1/r1−1/r2) = W[r1,r2]
Electric potential for each point of an electric field fully defines this field. If we know the electric potential in each point of a field, we don't have to know what kind of an object is the source of the field, nor its charge, nor shape.
To find the amount of work needed to move a charge q from a point in the electric field with a potential V1 to a point with potential V2 we use the formula W = q·(V2−V1)