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It seems that when something (magnetic field, electric field, current, etc.) changes, everything else changes. Can someone give me some hard and fast rules to get me started with this? Thank you.
Can someone explain induction and related concepts in a way that makes sense?
- Thread starter Med_Leviathan
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I can explain electrostatic induction.
Before I begin, remember that like charges repel (+ repels +, - repels -) and opposite charges attract (+ attracts -, vice versa).
Let's imagine that you have three metal bars (A, B and C)
Metal bar A is positively charged (more protons (+) than electrons (-))
Metal bar B and C have no charge (in other words, there are equal amounts of protons (+) and electrons (-))
Put the two bars (A and B) side to side like so: Bar A__Bar B (__ = two sides are near each other, but not touching)
Since A is positively charged, and it has more protons than electrons, electrons will be attracted to bar A. Therefore, electrons will accumulate on the left side of bar B (the side closer to bar A), creating a slight negative charge in the left side of bar B, which then creates a slight positive charge on the right side of bar B (the side away from bar A).
Diagram: ++BarA++ + Bar B = ++BarA++_-- Bar B ++
If you were to take away Bar A completely, then Bar B would return to a neutral charge. Why? Without the positive charge of Bar A, the electrons will no longer migrate to the left side, and the electrons will be dispersed evenly throughout bar B causing the entire bar to be neutrally charged.
Diagram: ++BarA++_-- BarB ++ - ++BarA++ = BarB
However, let's assume you take a third metal bar (Bar C) and place it next to the right side of Bar B (in this case, however, Bar B and Bar C do have contact.)
Diagram: ++BarA++_-- BarB++--BarC++
Remember that the right side of Bar B has a slightly positive charge because the electrons migrated from the right side to the left side. The slightly positive charge on the right side of Bar B will then create a slightly negative charge on the left side of bar C (the side nearest to bar B) the same way that Bar A created a negative charge on the left side of bar B (the side nearest bar A). However, since Bar B and Bar C actually touch, electrons from the negative side of bar c (closest to bar b) are transferred to Bar B [Note: Charge is always transferred by electrons. Since electrons orbit the nucleus, they can be transferred much easier than protons.} This means that Bar B gains electrons (thus gains a negative charge) and BarC loses electrons (thus gains a positive charge) even if you take away Bar A.
Diagram: BarB++--BarC.....--BarB-- _++BarC++
Another example can be found here:
http://en.wikipedia.org/wiki/Electrostatic_induction
Anyway, I tried to explain it as thoroughly as possible, but physics is hard to explain over the internet.
Hope it helps,
-Dr. P.
Before I begin, remember that like charges repel (+ repels +, - repels -) and opposite charges attract (+ attracts -, vice versa).
Let's imagine that you have three metal bars (A, B and C)
Metal bar A is positively charged (more protons (+) than electrons (-))
Metal bar B and C have no charge (in other words, there are equal amounts of protons (+) and electrons (-))
Put the two bars (A and B) side to side like so: Bar A__Bar B (__ = two sides are near each other, but not touching)
Since A is positively charged, and it has more protons than electrons, electrons will be attracted to bar A. Therefore, electrons will accumulate on the left side of bar B (the side closer to bar A), creating a slight negative charge in the left side of bar B, which then creates a slight positive charge on the right side of bar B (the side away from bar A).
Diagram: ++BarA++ + Bar B = ++BarA++_-- Bar B ++
If you were to take away Bar A completely, then Bar B would return to a neutral charge. Why? Without the positive charge of Bar A, the electrons will no longer migrate to the left side, and the electrons will be dispersed evenly throughout bar B causing the entire bar to be neutrally charged.
Diagram: ++BarA++_-- BarB ++ - ++BarA++ = BarB
However, let's assume you take a third metal bar (Bar C) and place it next to the right side of Bar B (in this case, however, Bar B and Bar C do have contact.)
Diagram: ++BarA++_-- BarB++--BarC++
Remember that the right side of Bar B has a slightly positive charge because the electrons migrated from the right side to the left side. The slightly positive charge on the right side of Bar B will then create a slightly negative charge on the left side of bar C (the side nearest to bar B) the same way that Bar A created a negative charge on the left side of bar B (the side nearest bar A). However, since Bar B and Bar C actually touch, electrons from the negative side of bar c (closest to bar b) are transferred to Bar B [Note: Charge is always transferred by electrons. Since electrons orbit the nucleus, they can be transferred much easier than protons.} This means that Bar B gains electrons (thus gains a negative charge) and BarC loses electrons (thus gains a positive charge) even if you take away Bar A.
Diagram: BarB++--BarC.....--BarB-- _++BarC++
Another example can be found here:
http://en.wikipedia.org/wiki/Electrostatic_induction
Anyway, I tried to explain it as thoroughly as possible, but physics is hard to explain over the internet.
Hope it helps,
-Dr. P.
