Electric Field Due To Point Charges - Physics Problems
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- เผยแพร่เมื่อ 6 ต.ค. 2024
- This video provides a basic introduction into the concept of electric fields. It explains how to calculate the magnitude and direction of an electric field created by a point charge as well as the net electric field produced by multiple point charges. It discusses the effect that an electric field has on positive and negative charges plus so much more. It contains plenty of examples and practice problems.
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Full 87 Minute Video - Part 1:
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Full 87 Minute Video - Part 1: bit.ly/3ZfWyUU
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00:05 Electric fields can be calculated using the formula F/q.
02:42 The electric field created by a point charge can be calculated using an equation.
08:22 The direction of the electric field can be determined based on the charge.
11:01 The direction of the electric field is due south.
15:47 The mass of the charge is approximately 0.255
18:05 An electric field can suspend a positive charge with a specific mass.
22:30 The electric field created by a 40 microcoulomb point charge placed at the origin is 14.4 kN/C at a point 5 meters away.
24:36 The electric field at point p is 1472.88 to 1476.16
29:01 The electron accelerates in the opposite direction to the electric field
31:14 The final speed of the electron is approximately 8.39 × 10^6 meters per second
35:51 The net electric field is 18 million.
38:07 The net electric field at point b is larger due to the closer distance and larger charge magnitude of q2.
42:41 The magnitude of the new electric field will be greater than 100 newtons per coulomb.
45:06 Electric field doubles when the magnitude of the charge doubles.
49:44 The net electric field at point A is not zero, but it is due to the combination of electric fields created by Q1 and Q2
51:46 The net electric field at point c will be directed east.
56:17 The net electric field is zero and e1 is equal to e2.
58:32 The exact location along the x-axis where the net electric field is zero can be calculated.
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at 48:16 for part b) where the distance is reduced by a factor of 3. He writes the distance as being 2/3 m but when entered into the equation he puts r as 1/3. So was the distanced reduced by a third, or is the distance a third. The wording is just confusing.
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In exercise six if it had no point of interset between those two charges how the eletric firld formula will be?
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Good lecture. Part b) assumes that the charged particle is almost touching the negative plate so that the distance traveled by the charge becomes 1cm. If not stated then distance moved by the charge can be anything less than 1cm.
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59:36
the answer of the net electric field is -12*10^5 N/C directed North
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Two parallel plates have a potential difference of 80 V. (a) A proton is shot from the negative plate towards the positive plate with an initial energy of 100 eV. What is the kinetic energy of the proton just before it strikes the positive plate? (b) Repeat if the proton is shot from the positive plate towards the negative plate.
I have a small question about the first practice question, if the electric field is directed to south, shouldn’t we put a negative sign so it becomes -5*10^6 ?
No
@@wizzyogwanz2750 why
Because it is asking for magnitude so you don't need to. Otherwise you are right it would be negative
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