The Electric Field "in front of" and "in back of" Two Point Charges

G.R.Dixon, September 19, 2008

The general formula for the electric field of point charge q is

(1)

In this formula e is a unit vector from the charges retarded position to the field evaluation point. And r is the scalar distance from the retarded position to the field evaluation point.

In the present case we consider two point charges, q1 and q2, whose common acceleration is constant, and whose positions are

, (2a)

. (2b)

(t is limited to values such that a|t|<c. Prior to t=0 the charges are moving in the negative x direction. At t=0 they come to rest at x=0 and x=r respectively. At all t>0 they move in the positive x direction.)

Since the time derivatives of e are zero in Eq. 1, we may simplify that formula to

. (3)

Or, defining E1 to be q1s field at the momentary position of q2 and E2 to be q2s field at the momentary position of q1, we have

, (4a)

. (4b)

Fig. 1 depicts the situation at some t<0. Note that r1<r and r2>r. Thus

. (5)

Figure 1

r1 and r2 at some t<0

The formulas for r1, dr1/dt, r2 and dr2/dt are

, (6a)

, (6b)

, (6c)

. (6d)

(These formulas are derived in Appendix A at the end of the article.) Plugging these values into Eqs. 4a and 4b produces E1 and E2.

Now when v is constant, the formulas for E1 and E2 are considerably simpler:

, (7a)

. (7b)

An interesting question is whether the values of E1 and E2 in the accelerated case equal those in the constant-velocity case. The question is readily answered by plotting the values for the two cases over a range of velocity components, say

. (8)

(A Visual Basic program that does this is provided in Appendix B.) Fig. 2 plots the accelerated values for E1 and E2 and the constant-velocity values. (The top curve is an overlay of the fields for q1; the bottom curve is an overlay of the fields for q2.) Note the perfect overlays. Evidently the "forward" and "backward" looking electric fields of a charge moving with constant velocity are the same as those of a charge subjected to a constant acceleration.

Figure 2

E1 and E2, constant v and constant a cases

***Appendix A***

In this appendix the formula for E1 is derived for any time t<0. (q1 and q2 are moving in the negative x direction, and are being accelerated in the positive x direction.) The cases for q2, and for t>0, are similarly derived.

Fig. A1 depicts the retarded position, etc., for q1.

Figure A1

r1, etc.

The retarded time is defined to be the present time minus a positive Dt.

. (A1)

The retarded velocity component is the constant acceleration times the retarded time.

. (A2)

Dx is the (positive) distance between the present position and the retarded position.

, (A3)

. (A4)

But

(see Fig. A1), (A5)

And by definition,

. (A6)

Subtracting A6 from A5:

. (A7)

And substituting from A4 produces

. (A8)

Solving A8 for Dt1 produces

. (A9)

And, since

, (A10)

differentiation of A9 produces

. (A11)

Finally,

(A12)

and

. (A13)

These can now be plugged into the formula for E1:

. (A14)

***Appendix B***

The Visual Basic program used to compute the fields follows.

Private Sub cmdConAccelE_Click()

Const N As Long = 1000

Const c As Double = 300000000# 'speed of light

Const vinit As Double = -0.99 * c

Const vfinal As Double = -0

Const dv As Double = (vfinal - vinit) / N

Const r As Double = 1 'separation of the charges

Const a As Double = 1000000 'constant acceleration

Const q As Double = 1 'magnitude of charges

Const eps0 As Double = 0.00000000000885

Const pi As Double = 3.141592654

Dim index As Long

Dim v(2 * N) As Double

Dim E1(2 * N), E2(2 * N), E(2 * N), E1p(2 * N), E2p(2 * N) As Double

Dim dt1, dt2, ddt1dt, ddt2dt As Double 'retarded time

Dim rp1, rp2, drp1dt, drp2dt As Double 'distance from retarded position to field eval pt

Open "c:/WINMCADC/Physics/TwoQ.PRN" For Output As #1

For index = 0 To N - 1

'For t<0

Debug.Print index

'and for each velocity considered,

v(index) = vinit + index * dv

'compute the constant-velocity fields.

E(index) = q / (4 * pi * eps0 * r ^ 2) * (1 - v(index) ^ 2 / c ^ 2)

E1p(index) = E(index)

E2p(index) = -E(index)

'Then compute the constant acceleration fields.

dt1 = -((c - v(index)) / a) + Sqr(((c - v(index)) / a) ^ 2 + 2 * r / a)

ddt1dt = 1 - c / (a * Sqr(((c - v(index)) / a) ^ 2 + 2 * r / a))

rp1 = c * dt1

drp1dt = c * ddt1dt

E1(index) = q / (4 * pi * eps0 * rp1 ^ 2) * (1 - 2 / c * drp1dt)

dt2 = (c + v(index)) / a - Sqr(((c + v(index)) / a) ^ 2 - 2 * r / a)

ddt2dt = 1 - c / (a * Sqr(((c + v(index)) / a) ^ 2 - 2 * r / a))

rp2 = c * dt2

drp2dt = c * ddt2dt

E2(index) = -q / (4 * pi * eps0 * rp2 ^ 2) * (1 - 2 / c * drp2dt)

Write #1, v(index), E1(index), E1p(index), E2(index), E2p(index)

Next index

For index = N To 2 * N - 1

'Do the same thing for times t>0.

Debug.Print index

v(index) = vfinal + (index - N) * dv

E(index) = q / (4 * pi * eps0 * r ^ 2) * (1 - v(index) ^ 2 / c ^ 2)

E1p(index) = E(index)

E2p(index) = -E(index)

dt1 = -((c - v(index)) / a) + Sqr(((c - v(index)) / a) ^ 2 + 2 * r / a)

ddt1dt = 1 - c / (a * Sqr(((c - v(index)) / a) ^ 2 + 2 * r / a))

rp1 = c * dt1

drp1dt = c * ddt1dt

E1(index) = q / (4 * pi * eps0 * rp1 ^ 2) * (1 - 2 / c * drp1dt)

dt2 = (c + v(index)) / a - Sqr(((c + v(index)) / a) ^ 2 - 2 * r / a)

ddt2dt = 1 - c / (a * Sqr(((c + v(index)) / a) ^ 2 - 2 * r / a))

rp2 = c * dt2

drp2dt = c * ddt2dt

E2(index) = -q / (4 * pi * eps0 * rp2 ^ 2) * (1 - 2 / c * drp2dt)

'Output all of the field values for plotting purposes.

Write #1, v(index), E1(index), E1p(index), E2(index), E2p(index)

Next index

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