Static Electric Fields

(Currently only functional on desktop)


Overview: The system shows an electric field generated by two charged particles. The change in electrostatic potential (voltage) that is generated by moving a particle from one point to another, while being acted on by the electric field, exists as the path integral from particle 1 to particle 2 over the electric field. As the electric field is conservative, the line integral is path independent and only changes when the charge (Q1 or Q2) changes or the position within the field changes. The line integral does not change when you simply alter the path alone.

Interactivity: You can change the charge of the particle to any value between [-200,200] using the black control panel at the top right. You can change the path in between the two charged particles by clicking and dragging the two small black dots. To move the particles, click and drag them to the desired location.



(Note, the potential energy calculation omits the constant \(\frac{1}{4 \pi \epsilon_0}\) where \(\epsilon_0\) is the Permittivity of the Vacuum.)

See below for some of the calculus and physics involved.




Math Source: Thomas Calculus 13th Edition Early Transcendentals (Thomas, Weir, Hass, Heil)

Line Integrals and Vector Fields

Line Integrals

  • Let \(f(x,y,z)\) be a real valued. Our goal is to integrate \(f\) over a curve \(C\) within the domain of \(f\), parametrized by \(r(t) = g(t)i + h(t)j + k(t)k \) with \(a \leq t \leq b \).
  • The values of \(f\) along \(C\) are given by the composite function \(f(g(t), h(t), k(t)) \). The goal is to integrate the composite function with respect to the arc length from \(t=a\) to \(t=b\).
  • Partition the curve \(C\) into \(n\) sub-arcs. An arbitrary sub-arc has length \(\Delta s_k\). Within each sub-arc, choose a point (\(x_k, y_k, z_k\)), evaluate \(f\) at that point,
    then form the sum \(S_n = \sum_{k=1}^n f(x_k, y_k, z_k) \Delta s_k \)
  • \(S_n\) is dependent upon both how we choose to partition our curve \(C\) along with the specific point (\(x_k, y_k, z_k\)), within the \(kth\) sub-arc, we choose to evaluate \(f\) at.
  • If \(f\) is continuous and its component functions \(g,h,k\) all posses continuous first derivatives then the sums \(S_n\) approach a limiting value as \(n \rightarrow \infty \) and the lengths of the sub-arcs \( \Delta s_k \) approach 0.

Definition If \(f\) is defined on a curve \(C\) given parametrically by \(r(t) = g(t)i + h(t)j + k(t)k \) with \(a \leq t \leq b \), then, provided the limit exists, the line integral of \(f\) over a curve \(C\) is \[\int_{C} f(x, y, z) ds = \lim_{n \to \infty} \sum_{k=1}^n f(x_k, y_k, z_k) \Delta s_k\]

If \(C\) is smooth for \(a \leq t \leq b \) (the rate of change in position of the curve as \(t\) varies \(\frac{dr}{dt} = v \) is continuous and never 0) then the limit in the above equation can be shown to exist.

Applying the Fundamental Theorem of Calculus to differentiate the arc length equation, with base point\(t_0 = a\): \[s(t) = \int_{a}^{t} |v( \tau )| d \tau \] expressing \(ds\) as \(ds = |v(t)| dt\) the line integral of \(f\) over \(C\) becomes \[\int_{C} f(x, y, z) ds = \int_{a}^{b} f(g(t), h(t), k(t)) |v(t)| dt \]

Vector Fields

A vector field, generally, is a function that assigns a vector to each point in its domain. Possibly being expressed as follows: \[F(x,y,z) = M(x,y,z)i + N(x,y,z)j + P(x,y,z)k\] The field is continuous if each component \(M\), \(N\), & \(P\) are continuos and differentiable if each component is differentiable.

Definition The gradient of a function \(f(x,y)\) at a point \(P_0(x_0,y_0)\) is the vector obtained by evaluating the partial derivatives of \(f\) at a point \(P_0\): \[ \nabla f = \frac{\delta f}{\delta x} i + \frac{ \delta f}{ \delta y} j\]

Definition The gradient field of a differentiable function \(f(x,y,z)\) is the field of gradient vectors \( \nabla f = \frac{\delta f}{\delta x} i + \frac{ \delta f}{ \delta y} j + \frac{ \delta f}{ \delta z} j\) where at each point \(x,y,z\) the gradient field gives a vector pointing in the direction of greatest change of \(f\) at a point. The magnitude of of the gradient being the value of the directional derivative at that point.

Line Integrals of Vector Fields

Intuitively, the line integral of a vector field along a curve \(C\) can be seen as the scalar tangential component of \(F\) along \(C\). The tangential component is given by the dot product \(F \cdot T = F \cdot \frac{dr}{ds}\)

Definition Let \(F\) be a vector field with continuous components defined along a smooth curve \(C\), where \(C\) is parametrized by \(r(t)\), \(a \leq t \leq b \). The line integral of \(F\) evaluated along \(C\) is: \[ \int_{C} F \cdot T ds = \int_{C} (F \cdot \frac{dr}{ds}) ds = \int_{C} F \cdot dr = \int_{a}^{b} F(r(t)) \cdot \frac{dr}{dt} dt \]

Work Done by a Force over a Curve in Space

Let \(F(x,y,z) = M(x,y,z)i + N(x,y,z)j + P(x,y,z)k\) and \(r(t) = g(t)i + h(t)j + k(t)k \) with \(a \leq t \leq b \) represents a smooth curve in space. We define the work done by a continuous force field \(F\) to move an object along \(C\) from a point \(A\) to another point \(B\) as follows:

  • Divide \(C\) into \(n\) sub-arcs \(P_{k-1},P_k\) each with length \( \Delta s_k \), going from \(A\) to \(B\).
  • Choose a point (\(x_k,y_k,z_k\)) in the sub-arc \(P_{k-1},P_k\) and let \(T(x_k,y_k,z_k)\) be the unit tangent vector at that chosen point.

The work \(W_k\) done to move an object along the sub-arc \(P_{k-1},P_k\) is approximated by the tangential component of the force \(F(x_k,y_k,z_k)\) multiplied by the arclength \( \Delta s_k\). \( \Delta s_k\) approximates the distance the object moves along a sub-arc. The total work done in moving an object from \(A\) to \(B\) is then approximated by summing the work done along each of the sub-arcs: \[W \approx \sum_{k=1}^n W_k \approx \sum_{k=1}^n F(x_k, y_k, z_k) \cdot T(x_k,y_k,z_k) \Delta s_k\]

For any subdivision of \(C\) into n subarcs and for any choice of points (\(x_k,y_k,z_k\)) within each subarc, as \(n \rightarrow \infty \) and \( \Delta s_k \rightarrow 0 \) the work sums approach a limiting value which is the line integral \[\int_{C} F \cdot T ds \]

Definition Let \(C\) be a smooth curve parametrized by \(r(t)\), \(a \leq t \leq b \), and \(F\) be a continuous force field over a region containing \(C\). The work done in moving an object from a point \(A = (r(a)\) to a point \(B = (r(b)\) along \(C\) is \[ W = \int_{C} F \cdot T ds = \int_{a}^{b} F(r(t)) \cdot \frac{dr}{dt} dt \]

Path Independence

If \(A\) and \(B\) are two points in an open region \(D\) in space, the line integral of \(F\) along a curve \(C\) for \(A\) to \(B\) for a field \(F\) defined on the region \(D\) is usually dependent on the path \(C\) taken. For certain fields though, the integral's value is the sm=ame for all paths taken from \(A\) to \(B\).

Definition: Let \(F\) be a vector field defined on an open region \(D\) in space. Suppose that for any two points \(A\) \(B\) in \(D\) the line integral \( \int_{C} F \cdot dr\) along a path \(C\) from \(A\) to \(B\) in \(D\) is the same over all paths from \(A\) to \(B\). Then the line integral \(\int_{C} F \cdot dr \) is Path Independent in D and the field F is said to be Conservative on D. Conservative comes from physics, referring to fields where the conservation of energy holds.

Definition If \(F\) is a vector field defined on \(D\) and \(F = \nabla f\) for some scalar function \(f\) on \(D\), then \(f\) is called a Potential Function for \(F\).

A field \(F\) is conservative if and only if it is the gradient field of some scalar function \(f\). A gravitational potential is a scalar function whose gradient field is a gravitational field, an electric potential is a scalar function whose gradient field is an electric field.

Once a potential function \(f\) (scalar) has been found for a vector field \(F\) we can evaluate the line integrals in the domain of \(F\) over any path between \(A\) and \(B\) by: \[ \int_{A}^{B} F \cdot dr = \int_{B}^{A} \nabla f \cdot dr = f(B)-f(A) \]



Physics Source: The Physics of Energy (Jaffe, Taylor)


Electric Fields, Forces, and Potential

Coulomb's Law: Two charges exert forces on one another that are equal in magnitude but opposite in direction. \[ F_{12} = - F_{21} = \frac{1}{4 \pi \epsilon_0} \frac{Q_1 Q_2}{r^2_{12}} \hat{r}_{12} \]

  • \(Q_1,Q_2\) are quantities of electric charge, \(r_{12}\) is the vector separating the two charges, \(F_{12}\) is the force exerted by the charge particle \(Q_1\) onto the charge particle \(Q_2\).
  • in SI units(The International System of Units) charge is measured in coulombs(C). The constant \(\epsilon_0\) is the Permittivity of the Vacuum and \(\frac{1}{4 \pi \epsilon_0} = 8.988 \times 10^9 Nm^2/C^2\)
  • The force between two coulombs of charge separated by one meter is ~ \(9 \times 10^9 N\)

For forces between electrically charges particles, the local mechanism that mediates the force is the Electric Field (for moving charges, Magnetic Fields are involved). An electric field is described by a vector \(E(x,t)\) at every point in space and time. The module restricts the focus to time-independent electric fields \(E(x)\)

A system of charged particles gives rise to an electric field \(E\). The field in turn exerts a force \(F\) onto any other charge \(q\). \[F = qE\]

The electric field produced at a point \(x\) in space by a single stationary charge \(Q\) at the point \(x_0\) is given by: \[E = \frac{Q}{4 \pi \epsilon_0 r^2} \hat{r}\] where \(r = x - x_0\) is the distance from the stationary charge \(Q\) to a point \(x\)

Superposition Principle The electric field produced by a set of charges is the vector sum, superposition, of the fields produced by each charge individually. For charges \(Q_i\) at positions \(x_i\) the electric field at position \(x\) is: \[E(x) = \frac{1}{4 \pi \epsilon_0} \sum_{i} Q_i \frac{x-x_i}{|x-x_i|^3}\]

If work is done by moving a charged particle against the force exerted by an electric field, the work is stored as Electrostatic Potential Energy. If we take a charged particle \(q\) and move it from positions \(x_1\) to \(x_2\) along a path \(P\) in the presence an electric field which is generated by a static charge distribution, the corresponding work done is: \[ W(P) = - \int_{P} dx \cdot F = - q \int_{P} dx \cdot E \]

For any electric field that is generated by a stationary charge distribution the integral depends only on the end points and not on the choice of path from positions \(x_1\) to \(x_2\). Electromagnetic Potential Energy is defined in units of energy per unit charge (voltage). The difference in electromagnetic energy potential energy when the charged particle is moved from positions \(x_1\) to \(x_2\) is, as above: \[ \Delta E_{EM} = - q \int_{x_1}^{x_2} dx \cdot E \]

If we divide the \(q\) from the above equation we obtain the difference in electrostatic potential \(V(x)\) (voltage) between positions \(x_1\) to \(x_2\). \[ V(x_2) - V(x_1) = \Delta E_{EM} / q = - \int_{x_1}^{x_2} dx \cdot E \]

The electric field is the negative of the gradient of \(V(x)\): \(E(x) = - \nabla V(x)\). The force on a charge is \(F = qE = -q \nabla V(x)\). Positive charges thus accelerate in the direction of decreasing potential \(V\) while negative charges accelerate in teh direction of increasing \(V\).

Electric Fields, Forces, and Potential

If a charge \(q\) is moved radially inward from \(r_1\) to \(r_2 \lt r_1 \) in the electric field of a point charge \(Q\), the work done is: \[ W(r_1,r_2) = \Delta E_{EM} =\] \[- \int_{r_1}^{r_2} dr \frac{Q}{4 \pi \epsilon_0 r^2} = \frac{qQ}{4 \pi \epsilon_0 r^2}(\frac{1}{r_2} - \frac{1}{r_1}) = q(V(r_2) - V(r_1))\] The electrostatic potential of a point charge is \(V(r) = \frac{Q}{4 \pi \epsilon_0 r} \). Moving a positive charge against the direction that an electric field points requires doing work on the charge, thus raising its potential.