The images below are visualizations of the electric charges and fields in a simple resistor-capacitor circuit with a switch. The resistor is a uniform copper wire joining the two plates. The switch is a small section of wire that initially is insulating, and becomes conducting at t=0. The different images represent computational steps in the evolution of the circuit.
The wires and plates are divided into computational cells, each a cube 0.5 mm on a side. The entire circuit is 25 mm (about one inches) across, the wires are 5 mm thick, and this image is a slice through the mid-plane of the circuit.
The time-step used in the calculations was 0.75 ps, which is about one-half the time for light to cross a computational cell. Light would take about 125 ps, or about 165 time steps, to cross the circuit.
The colors represent the amount of excess charge in each cell, from red (5000 or more positive elementary charges), to white (neutral), to blue (5000 or more negative elementary charges). The arrows show the magnitude and direction of the retarded electric field due to all the charges calculated at that point. (The very large electric fields present in and between the plates are not drawn so the much smaller fields in the wires are visible.)
Click on an image for a larger, clearer picture (725x560 pixels, about 50k each). There are small (10.5Mb) and large (22.5Mb) movies of the whole circuit, and small and large movies of the region around the switch.
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This shows the capacitor at t=0, when the insulating strip in the middle of the bottom wire becomes conductive (the switch is closed). Prior to this, the circuit was equilibrated for 1500 steps with the switch open, so the electric field is almost zero everywhere inside the wires. Note the large number of charges that have piled up on either side of the switch, resembling another finite capacitor. The right-hand frame is an enlargement of the bottom wire. |
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(5 steps = 3.75 ps) The positive
and negative charges that had piled up on either side of the
switch gap are starting to move towards each other and
neutralize. This will affect the electric field throughout the
circuit.
The electric field in nearby wires is the sum of the E-field from the switch charges and distant charges. The electric field from the switch charges resembles that of a finite capacitor, and the distant charges must be creating an electric field equal and opposite to the switch charges. As the switch charges neutralize, the electric field of the distant charges is exposed. |
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(10 steps = 7.5 ps) |
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(20 steps = 15 ps) We now start to see, in the region around the switch, the electric field produced by the distant charges. The charges in the interior of the wires are caused by the rapidly changing electric fields pushing charges more strongly near the switch than distant from the switch, resulting in a temporary accumulation of charges. As the circuit equilibrates, these volume charges disappear. |
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(30 steps = 22.5 ps) |
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(40 steps = 30 ps) The charges that had piled up at the borders of the switch have now been almost completely neutralized. We can now watch the changing charge distribution affected distant charges and slowly bring the circuit to a steady-state current flow. |
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(70 steps = 52.50 ps) |
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(150 steps = 112.5 ps) |
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(300 steps = 225 ps) The upper part of the circuit is now beginning to respond to the closing of the switch. It will take another two or three light-crossing times before the current reaches steady-state. |
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(800 steps = 600 ps) |
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(1500 steps = 1125 ps) |