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File:VFPt metal balls largesmall.svg

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Description
English: Electric field around a large and a small conducting sphere at opposite electric potential. The shape of the field lines is computed exactly, using the method of image charges with an infinite series of charges inside the two spheres. Field lines are always orthogonal to the surface of each sphere. In reality, the field is created by a continuous charge distribution at the surface of each sphere, indicated by small plus and minus signs.
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Author Geek3
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Source code
InfoField

SVG code

# paste this code at the end of VectorFieldPlot 1.10
# https://commons.wikimedia.org/wiki/User:Geek3/VectorFieldPlot
u = 100.0
doc = FieldplotDocument('VFPt_metal_balls_largesmall',
    commons=True, width=800, height=600, center=[400, 300], unit=u)

# define two spheres with position, radius and charge
s1 = {'p':sc.array([-1.0, 0.]), 'r':1.5}
s2 = {'p':sc.array([2.0, 0.]), 'r':0.5}

# make charge proportional to capacitance, which is proportional to radius.
s1['q'] = s1['r']
s2['q'] = -s2['r']
d = vabs(s2['p'] - s1['p'])
v12 = (s2['p'] - s1['p']) / d

# compute series of charges https://dx.doi.org/10.2174/1874183500902010032
charges = [[s1['p'][0], s1['p'][1], s1['q']], [s2['p'][0], s2['p'][1], s2['q']]]
r1 = r2 = 0.
q1, q2 = s1['q'], s2['q']
q0 = max(fabs(q1), fabs(q2))
for i in range(10):
    q1, q2 = -s1['r'] * q2 / (d - r2), -s2['r'] * q1 / (d - r1), 
    r1, r2 = s1['r']**2 / (d - r2), s2['r']**2 / (d - r1)
    p1, p2 = s1['p'] + r1 * v12, s2['p'] - r2 * v12
    charges.append([p1[0], p1[1], q1])
    charges.append([p2[0], p2[1], q2])
    if max(fabs(q1), fabs(q2)) < 1e-3 * q0:
        break

field = Field({'monopoles':charges})

# draw symbols
for c in charges:
    doc.draw_charges(Field({'monopoles':[c]}), scale=0.6*sqrt(fabs(c[2])))

gradr = doc.draw_object('linearGradient', {'id':'rod_shade', 'x1':0, 'x2':0,
    'y1':0, 'y2':1, 'gradientUnits':'objectBoundingBox'}, group=doc.defs)
for col, of in (('#666', 0), ('#ddd', 0.6), ('#fff', 0.7), ('#ccc', 0.75),
    ('#888', 1)):
    doc.draw_object('stop', {'offset':of, 'stop-color':col}, group=gradr)
gradb = doc.draw_object('radialGradient', {'id':'metal_spot', 'cx':'0.53',
    'cy':'0.54', 'r':'0.55', 'fx':'0.65', 'fy':'0.7',
    'gradientUnits':'objectBoundingBox'}, group=doc.defs)
for col, of in (('#fff', 0), ('#e7e7e7', 0.15), ('#ddd', 0.25),
    ('#aaa', 0.7), ('#888', 0.9), ('#666', 1)):
    doc.draw_object('stop', {'offset':of, 'stop-color':col}, group=gradb)

ball_charges = []
for ib in range(2):
    ball = doc.draw_object('g', {'id':'metal_ball{:}'.format(ib+1),
        'transform':'translate({:.3f},{:.3f})'.format(*([s1, s2][ib]['p'])),
        'style':'fill:none; stroke:#000;stroke-linecap:square', 'opacity':1})
    
    # draw rods
    if ib == 0:
        x1, x2 = -4.1 - s1['p'][0], -0.9 * s1['r']
    else:
        x1, x2 = 0.9 * s2['r'], 4.1 - s2['p'][0]
    doc.draw_object('rect', {'x':x1, 'width':x2-x1,
        'y':-0.1/1.2+0.01, 'height':0.2/1.2-0.02,
        'style':'fill:url(#rod_shade); stroke-width:0.02'}, group=ball)
    
    # draw metal balls
    doc.draw_object('circle', {'cx':0, 'cy':0, 'r':[s1, s2][ib]['r'],
        'style':'fill:url(#metal_spot); stroke-width:0.02'}, group=ball)
    ball_charges.append(doc.draw_object('g',
        {'style':'stroke-width:0.02'}, group=ball))

# find well-distributed start positions of field lines
def get_startpoint_function(startpath, field):
    '''
    Given a vector function startpath(t), this will return a new
    function such that the scalar parameter t in [0,1] progresses
    indirectly proportional to the orthogonal field strength.
    '''
    def dstartpath(t):
        return (startpath(t+1e-6) - startpath(t-1e-6)) / 2e-6
    def FieldSum(t0, t1):
        return ig.quad(lambda t: sc.absolute(sc.cross(
            field.F(startpath(t)), dstartpath(t))), t0, t1)[0]
    Ftotal = FieldSum(0, 1)
    def startpos(s):
        t = op.brentq(lambda t: FieldSum(0, t) / Ftotal - s, 0, 1)
        return startpath(t)
    return startpos

startp = []
def startpath1(t):
    phi = 2. * pi * t
    return (sc.array(s2['p']) + 1.5 * sc.array([cos(phi), sin(phi)]))
start_func1 = get_startpoint_function(startpath1, field)
nlines1 = 16
for i in range(nlines1):
    startp.append(start_func1((0.5 + i) / nlines1))

def startpath2(t):
    phi = 2. * pi * (0.195 + 0.61 * t)
    return (sc.array(s1['p']) + 1.5 * sc.array([cos(phi), -sin(phi)]))
start_func2 = get_startpoint_function(startpath2, field)
nlines2 = 14
for i in range(nlines2):
    startp.append(start_func2((0.5 + i) / nlines2))

# draw the field lines
for p0 in startp:
    line = FieldLine(field, p0, directions='both', maxr=7.)
    
    # draw little charge signs near the surface
    path_minus = 'M {0:.5f},0 h {1:.5f}'.format(-2./u, 4./u)
    path_plus = 'M {0:.5f},0 h {1:.5f} M 0,{0:.5f} v {1:.5f}'.format(-2./u, 4./u)
    for si in range(2):
        sphere = [s1, s2][si]
        
        # check if fieldline ends inside the sphere
        for ci in range(2):
            if vabs(line.get_position(ci) - sphere['p']) < sphere['r']:
                # find the point where the field line cuts the surface
                t = op.brentq(lambda t: vabs(line.get_position(t)
                    - sphere['p']) - sphere['r'], 0., 1.)
                pr = line.get_position(t) - sphere['p']
                cpos = 0.9 * sphere['r'] * pr / vabs(pr)
                doc.draw_object('path', {'stroke':'black', 'd':
                    [path_plus, path_minus][ci],
                    'transform':'translate({:.5f},{:.5f})'.format(
                        round(u*cpos[0])/u, round(u*cpos[1])/u)},
                        group=ball_charges[si])
    
    arrow_d = 2.0
    of = [0.5 + s1['r'] / arrow_d, 0.5, 0.5, 0.5 + s2['r'] / arrow_d]
    doc.draw_line(line, arrows_style={'dist':arrow_d, 'offsets':of})
doc.write()

Licensing

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I, the copyright holder of this work, hereby publish it under the following license:
w:en:Creative Commons
attribution share alike
This file is licensed under the Creative Commons Attribution-Share Alike 4.0 International license.
You are free:
  • to share – to copy, distribute and transmit the work
  • to remix – to adapt the work
Under the following conditions:
  • attribution – You must give appropriate credit, provide a link to the license, and indicate if changes were made. You may do so in any reasonable manner, but not in any way that suggests the licensor endorses you or your use.
  • share alike – If you remix, transform, or build upon the material, you must distribute your contributions under the same or compatible license as the original.

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Date/TimeThumbnailDimensionsUserComment
current20:05, 30 December 2018Thumbnail for version as of 20:05, 30 December 2018800 × 600 (46 KB)Geek3 (talk | contribs)User created page with UploadWizard

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