Ch 22: Electric Force & Field; Gauss' LawSee all chapters

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Electric Charge | 15 mins | 0 completed | Learn |

Charging Objects | 7 mins | 0 completed | Learn |

Charging By Induction | 4 mins | 0 completed | Learn |

Conservation of Charge | 8 mins | 0 completed | Learn |

Coulomb's Law (Electric Force) | 56 mins | 0 completed | Learn Summary |

Electric Field | 46 mins | 0 completed | Learn Summary |

Parallel Plate Capacitors | 16 mins | 0 completed | Learn |

Electric Field Lines | 13 mins | 0 completed | Learn |

Dipole Moment | 7 mins | 0 completed | Learn |

Electric Fields in Conductors | 5 mins | 0 completed | Learn |

Electric Flux | 19 mins | 0 completed | Learn Summary |

Gauss' Law | 24 mins | 0 completed | Learn Summary |

Concept #1: Intro to Capacitors

**Transcript**

Hey guys in this video we're going to be talking about things called capacitors and the electric fields produced by them. So, let's get to it, if we take two parallel plates of equal and opposite charge they produce a uniform electric field between them, okay? Let's say, this is plus q and this is minus q they're equal opposite charge, what does uniform mean, when I say a uniform electric field uniform means it's the same magnitude everywhere. So, if I were to measure the electric field here, call that v1 and I were to measure the electric field here and call that e2 and then measure the electric field here and call that e3 these would all be equal to one another. So, that's one half of uniform same everywhere the other half of uniform is it doesn't change with time, okay? So, it's constant in position, meaning wherever position I choose it's constant and it's constant in time, meaning that as time changes it doesn't change, okay? Now, what about the direction of this electric field? Well, if I were to choose an arbitrary point between them we would have two components to that electric field, we would have a component due to the positive charges on the positive plate and a component due to the negative charges on the negative plates, the positive plate is going to produce an electric field away from it which is down, the negative plate is going to produce an electric field towards it which is also down. So, everywhere between the two plates on the capacitors, the electric field is going to point from the positive plate to the negative plate and it's going to be constant. So, it's typically drawn like this, okay? So, if I take two plates, one a positive q, one of negative q, the electric field just look like that, okay? No big deal, capacitors they produce a uniform electric field and that electric field goes from positive place to the negative plate really simple, okay? Well, what is the electric field within the capacitor and what is the electric field outside the capacitor, I mean, what is the magnitude, okay? So, let's talk about that. Between the plates the magnitude is simply q, the charge on each plate divided by epsilon naught times A, where a is the area of each place epsilon naught is a new constant it's 8.85 times 10 to the negative 12 and the units which aren't important are F farad over m, okay? For. Now, don't worry about the unit's those will appear again later, okay? 8.85 times 10 to the negative 12 is the important number to remember. Now, this is a constant that you're most likely going to be given on an exam or in a homework problem. So, it's not essential to memorize it like we've done for the elementary charge and for Coulomb's constant, okay? And this constant epsilon naught is simply called the vacuum permittivity, okay? That name is also not super important but obviously it has to be named, constants are always named and that's what this one is called. Now, what about outside of the electric fields plates? Well, outside it's simply 0, very easy, no electric field outside of the capacitors, the electric field is contained entirely within the two capacitor plates. So, you have this uniform electric field inside and you have no electric field outside. Alright, let's do an example.

The field between two parallel plates is 1,000 Newtons per Coulomb, if the plates have an area of five square centimeters, what is the charge on each plate, okay? So, we're going to use the equation for the electric fields within a capacitor which is E equals q over epsilon naught A, we want to find Q. So, what we need to do is we need to take to this denominator and multiply it up into the numerator of the other side. that then tells us that q is epsilon naught, A times E, epsilon naught being a 8.85 times 10 to the negative 12. Remember, you don't have to memorize that. Now, 5 square centimeters one square centimeter is one 10 thousandth of a square meter, okay? So, this is going to be 5 times 10 to the negative 4 square meters and the electric field is 1,000 Newtons per Coulomb. So, what's the charge going to be? 4.4 three times 10 to the negative twelve coulombs just plug it into your calculator. Alright, very straightforward very simple application of our equation for the electric field of the within a capacitor, alright? Now, example two, a capacitor produces an electric field E, when it is formed by two plates that have charges q a negative q, what happens to the electric field, if the charge doubles but the area of the plates halves. So, E, remember, equals q divided by epsilon naught A, okay? So, if the q becomes twice as big then the electric field has to become twice as big, E is what we recall directly proportional to q, if a becomes of 1/2 then E actually doubles again E is what we call directly proportional to 1 over A. So, what that cumulative effect is, is double of double E which means E increases by 4, okay? Now, if you're uncomfortable these proportionalities there's a much more precise way of doing this, we can say that q becomes 2Q and A becomes 1/2 A, okay? So, the new electric field is just new Q divided by epsilon naught times new A, new Q is 2 times old Q divided by epsilon naught, new A is 1/2 times old A. So, this if I pull the 2 out in the 1/2 out, this is 2 over 1/2 times Q over epsilon naught A, this is the old electric field and 2 divided by 1/2 is actually 4. So, you see it's multiplied by 4 in the end, okay? So, this is the more precise way of doing it. Alright, that wraps up our discussion on capacitors.

Practice: An electron moves into a capacitor at an initial speed of 150 m/s. If the electron enters exactly halfway between the plates, how far will the electron move horizontally before it strikes one of the plates? Which plate will it strike?

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Concept #1: Intro to Capacitors

Practice #1: Kinematics in Capacitor

Consider the three infinite sheets of charge shown in the figure. The first sheet has chage density σ1 = +10μC/m2 and lies on the yz-plane. The second sheet has charge density of σ2 = +10μC/m2 and is parallel to the yz-plane at x = 0.10 m. The third sheet has charge density of σ3 = +10μC/m2 and is parallel to the yz-plane at x = 0.20 m. Determine the electric field at the point P (0.05 m, 0).

The two infinite, conducting plates of thickness = h and separation = d are oppositely charged. The charge per unit area on the top plate is σ , and the charge on the bottom plate is σ , where σ is a positive number. What is the magnitude of the electric field in each region depicted in the picture?

Four parallel infinite sheets of charge spaced by 4 cm between adjacent sheets are shown edgewise in the figure. Their charge densities are as indicated with σ = 2.0 μC/m2. What is the magnitude of the electric field at point P midway between two sheets as shown?
(1) 4.5 × 105 N/C
(2) 2.3 × 105 N/C
(3) 1.13 × 105 N/C
(4) 9.0 × 105 N/C
(5) 0

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