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EN_B00000_S02952_W000002 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S02952_W000003 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S02952_W000004 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S02962_W000000 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S02962_W000001 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S02962_W000004 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.3152,
"duration": 4.058,
"id": "EN_B00000_S02962_W000004",
"language": "en",
"speaker": "EN_B00000_S02962",
"text": " So first, we need to take the difference between the number of protons and electrons.",
"wav": "EN_B00000/EN_B00000_S02962/mp3/EN_B00000_S02962_W000004.mp3"
} | EN_B00000_S02962_W000004.mp3 | [
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EN_B00000_S02962_W000005 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.4415,
"duration": 6.062,
"id": "EN_B00000_S02962_W000005",
"language": "en",
"speaker": "EN_B00000_S02962",
"text": " And then we can multiply that difference by the charge of each particle.",
"wav": "EN_B00000/EN_B00000_S02962/mp3/EN_B00000_S02962_W000005.mp3"
} | EN_B00000_S02962_W000005.mp3 | [
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EN_B00000_S02962_W000006 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.4499,
"duration": 8.201,
"id": "EN_B00000_S02962_W000006",
"language": "en",
"speaker": "EN_B00000_S02962",
"text": " The charge of a proton is 1.6 times 10 to the negative 19 coulombs. For an electron, it's simply negative of that number.",
"wav": "EN_B00000/EN_B00000_S02962/mp3/EN_B00000_S... | EN_B00000_S02962_W000006.mp3 | [
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EN_B00000_S02962_W000007 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"duration": 17.589,
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"language": "en",
"speaker": "EN_B00000_S02962",
"text": " So the number of protons is 5 times 10 to the 13. The number of electrons is 8.5 times 10 to the 13. Now, because we have more electrons than protons, we should expect that ou... | EN_B00000_S02962_W000007.mp3 | [
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EN_B00000_S02962_W000008 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.3922,
"duration": 4.142,
"id": "EN_B00000_S02962_W000008",
"language": "en",
"speaker": "EN_B00000_S02962",
"text": " So 5 minus 8.5, that's gonna be negative 3.5.",
"wav": "EN_B00000/EN_B00000_S02962/mp3/EN_B00000_S02962_W000008.mp3"
} | EN_B00000_S02962_W000008.mp3 | [
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EN_B00000_S02962_W000009 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.1944,
"duration": 5.28,
"id": "EN_B00000_S02962_W000009",
"language": "en",
"speaker": "EN_B00000_S02962",
"text": " So let's take that number and multiply it by the charge of a proton.",
"wav": "EN_B00000/EN_B00000_S02962/mp3/EN_B00000_S02962_W000009.mp3"
} | EN_B00000_S02962_W000009.mp3 | [
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EN_B00000_S02962_W000010 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.3502,
"duration": 18.862,
"id": "EN_B00000_S02962_W000010",
"language": "en",
"speaker": "EN_B00000_S02962",
"text": " So the total charge on the dust particle is negative 5.6 times 10 to the negative 6 coulombs. The negative sign simply tells us that there's more electrons than protons in thi... | EN_B00000_S02962_W000010.mp3 | [
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EN_B00000_S02962_W000011 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.3956,
"duration": 6.265,
"id": "EN_B00000_S02962_W000011",
"language": "en",
"speaker": "EN_B00000_S02962",
"text": " One microcloom is one times ten to the negative six cooms. So this cancels.",
"wav": "EN_B00000/EN_B00000_S02962/mp3/EN_B00000_S02962_W000011.mp3"
} | EN_B00000_S02962_W000011.mp3 | [
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EN_B00000_S02962_W000012 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.3634,
"duration": 4.975,
"id": "EN_B00000_S02962_W000012",
"language": "en",
"speaker": "EN_B00000_S02962",
"text": " Thus our final answer is simply negative 5.6 microclooms.",
"wav": "EN_B00000/EN_B00000_S02962/mp3/EN_B00000_S02962_W000012.mp3"
} | EN_B00000_S02962_W000012.mp3 | [
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EN_B00000_S02962_W000013 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.4258,
"duration": 15.382,
"id": "EN_B00000_S02962_W000013",
"language": "en",
"speaker": "EN_B00000_S02962",
"text": " Calculate the magnitude and direction of the net electric force acting on the negative 40 microcoulomb point charge. So let's begin by drawing a picture.",
"wav": "EN_B00000... | EN_B00000_S02962_W000013.mp3 | [
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EN_B00000_S02962_W000014 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.3943,
"duration": 5.687,
"id": "EN_B00000_S02962_W000014",
"language": "en",
"speaker": "EN_B00000_S02962",
"text": " So we have a positive 50 microcalume charge.",
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EN_B00000_S02962_W000015 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " And east of that, we have a negative 40.",
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EN_B00000_S02962_W000016 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " Microcalume charge. And so that's going to be eight meters east of the 50 microcalume charge.",
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EN_B00000_S02962_W000017 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " Of the force acting on the second charge to the right. Let's call this Q1. And the charge on the right will call it Q2. So what is the net force acting on Q2? Now what do we k... | EN_B00000_S02962_W000017.mp3 | [
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EN_B00000_S02962_W000018 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " And that force is directed west. So we could eliminate answer choice A and answer choice D. So we have the direction. Now what we need to do is calculate the magnitude of the ... | EN_B00000_S02962_W000018.mp3 | [
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EN_B00000_S02962_W000019 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " And so we could use this formula to do it. It's kq1q2 over r squared. So k is 9 times 10 to the 9, and the units are newtons, square meters,",
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EN_B00000_S02962_W000020 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " And then we have q1, which is 50 microcoulombs. So keep in mind, a microcoulomb is 10 to the minus 6 coulombs.",
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EN_B00000_S02962_W000021 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S02962_W000022 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S02962_W000023 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S02962_W000024 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " And so the force is going to be 0.281 N.",
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EN_B00000_S02962_W000025 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"speaker": "EN_B00000_S02962",
"text": " Number three, calculate the magnitude of the electric field created by a negative 75 microcoulomb point charge at a distance of 50 centimeters east from the point charge. So l... | EN_B00000_S02962_W000025.mp3 | [
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EN_B00000_S02962_W000026 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " Now at a distance of 50 centimeters east, this is going to be the point of interest. So we want to calculate the electric field at that point. Now let's review a few things. L... | EN_B00000_S02962_W000026.mp3 | [
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EN_B00000_S02962_W000027 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"speaker": "EN_B00000_S02962",
"text": " The electric field points towards the negative point charge in all directions.",
"wav": "EN_B00000/EN_B00000_S02962/mp3/EN_B00000_S02962_W000027.mp3"
} | EN_B00000_S02962_W000027.mp3 | [
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EN_B00000_S02962_W000028 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"language": "en",
"speaker": "EN_B00000_S02962",
"text": " So what is the direction of the electric field at point P, created by Q1?",
"wav": "EN_B00000/EN_B00000_S02962/mp3/EN_B00000_S02962_W000028.mp3"
} | EN_B00000_S02962_W000028.mp3 | [
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EN_B00000_S02962_W000029 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"id": "EN_B00000_S02962_W000029",
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"speaker": "EN_B00000_S02962",
"text": " So the direction is going to be west.",
"wav": "EN_B00000/EN_B00000_S02962/mp3/EN_B00000_S02962_W000029.mp3"
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EN_B00000_S02962_W000030 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"speaker": "EN_B00000_S02962",
"text": " It's going to be directed towards Q1.",
"wav": "EN_B00000/EN_B00000_S02962/mp3/EN_B00000_S02962_W000030.mp3"
} | EN_B00000_S02962_W000030.mp3 | [
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EN_B00000_S02962_W000031 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"duration": 11.885,
"id": "EN_B00000_S02962_W000031",
"language": "en",
"speaker": "EN_B00000_S02962",
"text": " Now to calculate the magnitude of the electric field, it's going to be kq over r squared. So k is 9 times 10 to the 9th, newtons times square meters.",
"wav": "EN_B00000/EN_B... | EN_B00000_S02962_W000031.mp3 | [
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EN_B00000_S02962_W000032 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"speaker": "EN_B00000_S02962",
"text": " Divided by coloom squared, which I'm going to write as colooms times colooms. Q is 75 microcolooms.",
"wav": "EN_B00000/EN_B00000_S02962/mp3/EN_B00000_S02962_W000032.mp3"
} | EN_B00000_S02962_W000032.mp3 | [
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EN_B00000_S02962_W000033 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " Now because the electric field is a vector and we already know its direction, I'm not going to use negative 75. I'm going to make it positive. R is 50 centimeters. Now we need... | EN_B00000_S02962_W000033.mp3 | [
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EN_B00000_S02962_W000034 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " R is the distance between the point charge and the point of interest.",
"wav": "EN_B00000/EN_B00000_S02962/mp3/EN_B00000_S02962_W000034.mp3"
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EN_B00000_S02962_W000035 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"language": "en",
"speaker": "EN_B00000_S02962",
"text": " So right now we could cancel the unit square meters and we could cancel one of the column units. So the electric field is going to be.",
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EN_B00000_S02962_W000036 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.2252,
"duration": 7.318,
"id": "EN_B00000_S02962_W000036",
"language": "en",
"speaker": "EN_B00000_S02962",
"text": " 9 times 10 to the 9 times 75 times 10 to the minus 6 divided by .5 squared.",
"wav": "EN_B00000/EN_B00000_S02962/mp3/EN_B00000_S02962_W000036.mp3"
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EN_B00000_S02962_W000037 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " So it's about 2.7 million, which we could say is 2.7 times 10 to the 6 Newtons per Coulomb. So that's the unit for electric field.",
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EN_B00000_S02962_W000038 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.1085,
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"id": "EN_B00000_S02962_W000038",
"language": "en",
"speaker": "EN_B00000_S02962",
"text": " Number four, calculate the magnitude and direction of the electric force exerted on a negative 80 microcoulomb point charge by an electric field",
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EN_B00000_S02962_W000039 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"language": "en",
"speaker": "EN_B00000_S02962",
"text": " So let's say we have two charge metal plates.",
"wav": "EN_B00000/EN_B00000_S02962/mp3/EN_B00000_S02962_W000039.mp3"
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EN_B00000_S02962_W000040 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"duration": 3.209,
"id": "EN_B00000_S02962_W000040",
"language": "en",
"speaker": "EN_B00000_S02962",
"text": " We're gonna say the plate at the bottom is positively charged.",
"wav": "EN_B00000/EN_B00000_S02962/mp3/EN_B00000_S02962_W000040.mp3"
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EN_B00000_S02962_W000041 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"speaker": "EN_B00000_S02962",
"text": " And the plate above it is negatively charged.",
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EN_B00000_S02962_W000042 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"language": "en",
"speaker": "EN_B00000_S02962",
"text": " The electric field will emanate from the positive plate and travel towards the negative plate.",
"wav": "EN_B00000/EN_B00000_S02962/mp3/EN_B00000_S02962_W000042.mp3"
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EN_B00000_S02962_W000043 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"speaker": "EN_B00000_S02962",
"text": " Now let's place a negative charge in the electric field.",
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EN_B00000_S02962_W000044 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"speaker": "EN_B00000_S02962",
"text": " What is the direction of the electric force that this negative charge will feel? And let's contrast that with a positive charge.",
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EN_B00000_S02962_W000045 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"duration": 13.837,
"id": "EN_B00000_S02962_W000045",
"language": "en",
"speaker": "EN_B00000_S02962",
"text": " will fill a force that will accelerate it in the same direction as the electric field. A negative charge will fill an electric force that will accelerate it opposite to the di... | EN_B00000_S02962_W000045.mp3 | [
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EN_B00000_S02962_W000046 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.3752,
"duration": 7.267,
"id": "EN_B00000_S02962_W000046",
"language": "en",
"speaker": "EN_B00000_S02962",
"text": " So the positive charge will be attracted to the negatively charged plate, and the negative charge will be attracted to the positively charged plate.",
"wav": "EN_B00000/EN_B0... | EN_B00000_S02962_W000046.mp3 | [
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EN_B00000_S02962_W000047 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"duration": 3.039,
"id": "EN_B00000_S02962_W000047",
"language": "en",
"speaker": "EN_B00000_S02962",
"text": " But make sure you understand that. A positive charge",
"wav": "EN_B00000/EN_B00000_S02962/mp3/EN_B00000_S02962_W000047.mp3"
} | EN_B00000_S02962_W000047.mp3 | [
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EN_B00000_S02962_W000048 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.3666,
"duration": 29.542,
"id": "EN_B00000_S02962_W000048",
"language": "en",
"speaker": "EN_B00000_S02962",
"text": " We'll fill a force that is directed south and not north. So we could eliminate B and D. Now let's calculate the magnitude of the electric force. The electric force is the prod... | EN_B00000_S02962_W000048.mp3 | [
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EN_B00000_S02962_W000049 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.432,
"duration": 13.073,
"id": "EN_B00000_S02962_W000049",
"language": "en",
"speaker": "EN_B00000_S02962",
"text": " And the charge is 80 times 10 to the negative 6 coulombs. Once again, we're not going to worry about the negative sign. Notice that the unit coulombs cancel, giving us the unit... | EN_B00000_S02962_W000049.mp3 | [
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EN_B00000_S02962_W000050 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.3524,
"duration": 8.269,
"id": "EN_B00000_S02962_W000050",
"language": "en",
"speaker": "EN_B00000_S02962",
"text": " So what we have is 4 times 80, which is 320. 10 to the 5 times 10 to the negative 6.",
"wav": "EN_B00000/EN_B00000_S02962/mp3/EN_B00000_S02962_W000050.mp3"
} | EN_B00000_S02962_W000050.mp3 | [
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EN_B00000_S02962_W000051 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"duration": 8.268,
"id": "EN_B00000_S02962_W000051",
"language": "en",
"speaker": "EN_B00000_S02962",
"text": " You need to add the exponents, 5 plus negative 6 is negative 1. So basically this is 320 divided by 10, which is 32.",
"wav": "EN_B00000/EN_B00000_S02962/mp3/EN_B00000_S02962... | EN_B00000_S02962_W000051.mp3 | [
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EN_B00000_S02962_W000052 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.2765,
"duration": 7.844,
"id": "EN_B00000_S02962_W000052",
"language": "en",
"speaker": "EN_B00000_S02962",
"text": " So the answer is 32 newtons direct itself. So this corresponds to answer choice A.",
"wav": "EN_B00000/EN_B00000_S02962/mp3/EN_B00000_S02962_W000052.mp3"
} | EN_B00000_S02962_W000052.mp3 | [
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EN_B00000_S02962_W000053 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.1782,
"duration": 6.91,
"id": "EN_B00000_S02962_W000053",
"language": "en",
"speaker": "EN_B00000_S02962",
"text": " Number five. Which of the following statements is false? So let's analyze each one.",
"wav": "EN_B00000/EN_B00000_S02962/mp3/EN_B00000_S02962_W000053.mp3"
} | EN_B00000_S02962_W000053.mp3 | [
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EN_B00000_S02962_W000054 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
"dnsmos": 3.3963,
"duration": 12.377,
"id": "EN_B00000_S02962_W000054",
"language": "en",
"speaker": "EN_B00000_S02962",
"text": " Let's look at AFERS. Electric field is defined as the force exerted on a tiny positive test charge divided by the magnitude of that charge. Is that true or false?",
"wav": "... | EN_B00000_S02962_W000054.mp3 | [
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EN_B00000_S02962_W000055 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S02962_W000056 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S02962_W000057 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S02962_W000058 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " Is equal to one joule of potential energy per one column of charge.",
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EN_B00000_S02962_W000059 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S02962_W000060 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S02962_W000061 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " You can also calculate the electric field using this formula, particularly if you have two parallel plates, if they're both charged.",
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EN_B00000_S02962_W000062 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S02962_W000063 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S02962_W000064 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S02962_W000065 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"language": "en",
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"text": " Is not a unit of voltage, but it's a unit of energy. And that's what you need to watch out for. So here's the conversion.",
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EN_B00000_S02962_W000066 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " Times 10 to the negative 19 joules.",
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EN_B00000_S02962_W000067 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " Just as one electron has a charge of 1.602 times 10 to the negative 19 colubes.",
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EN_B00000_S02962_W000068 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S02962_W000069 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " So one electron has a charge of 1.6 times 10 to the negative 19 C.",
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EN_B00000_S02962_W000070 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S02962_W000071 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S02962_W000072 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S02962_W000073 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S02962_W000074 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " Volts is the unit of electric potential.",
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EN_B00000_S02962_W000075 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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EN_B00000_S02962_W000086 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"duration": 5.331,
"id": "EN_B00000_S02962_W000086",
"language": "en",
"speaker": "EN_B00000_S02962",
"text": " So for this particular problem, the final speed of the electron is going to be the square root",
"wav": "EN_B00000/EN_B00000_S02962/mp3/EN_B00000_S02962_W000086.mp3"
} | EN_B00000_S02962_W000086.mp3 | [
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EN_B00000_S02962_W000087 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"speaker": "EN_B00000_S02962",
"text": " of negative two times the charge of the electron times the voltage that",
"wav": "EN_B00000/EN_B00000_S02962/mp3/EN_B00000_S02962_W000087.mp3"
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EN_B00000_S02962_W000088 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"text": " It accelerates through, divided by the mass of the electron.",
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EN_B00000_S02962_W000089 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"language": "en",
"speaker": "EN_B00000_S02962",
"text": " We know it's negative 1.6 times 10 to the negative 19 coulombs.",
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} | EN_B00000_S02962_W000089.mp3 | [
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EN_B00000_S02962_W000090 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"duration": 3.447,
"id": "EN_B00000_S02962_W000090",
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"speaker": "EN_B00000_S02962",
"text": " The voltage is positive 3000 volts.",
"wav": "EN_B00000/EN_B00000_S02962/mp3/EN_B00000_S02962_W000090.mp3"
} | EN_B00000_S02962_W000090.mp3 | [
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EN_B00000_S02962_W000091 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"language": "en",
"speaker": "EN_B00000_S02962",
"text": " The mass of the electron, keep in mind that's still inside the square root, the mass is 9.11 times 10 to the negative 31 kilograms. So go ahead and type it in exactly the way ... | EN_B00000_S02962_W000091.mp3 | [
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EN_B00000_S02962_W000092 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"duration": 18.438,
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"language": "en",
"speaker": "EN_B00000_S02962",
"text": " So that is the approximate speed of the electron after it accelerates from us through this potential difference. So D is the correct answer. Number eight, a voltage of 120 vol... | EN_B00000_S02962_W000092.mp3 | [
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EN_B00000_S02962_W000093 | hf://datasets/amphion/Emilia-Dataset@bcaad00d13e7c101485990a46e88f5884ffed3fc/EN/EN-B000002.tar | {
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"language": "en",
"speaker": "EN_B00000_S02962",
"text": " Across two parallel plates separated by a distance of 10 centimeters. What is the magnitude of the electric field that exists between the two plates? So let's draw a picture."... | EN_B00000_S02962_W000093.mp3 | [
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