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All Physics 19 Tips
| Question | Answer |
|---|---|
| Are the electric field force and electric field scalar or vector? | Vector |
| Electric Potential and Potential Energy: | both are scalar quantities, the electric potential (Va) at point (a) is defined as the electric potential energy (EPEa) per unit charge (q) placed at that point. These revolve around formulas given on the sheet → Va = EPEa/q and EPEa = qVa |
| Electric Potential Due to a Point Charge: | the electric potential due to a point charge at a distance (r) from the charge is given by the following formula. Note that the zero potential is arbitrarily chosen to be at infinity. Revolves around a formula on the sheet → V = kq/r |
| Tip about Electric Potential: | Notice that there is NO absolute value for the charges here! |
| Units Associated with Electric Potential: | The SI unit of electric potential is the volt (V) where 1 V = 1 J/C. |
| How to determine the electric potential at the center of a square? | Figure out the electric potential for each charge at the center of the square and then add them up (formula for total potential given on sheet). Use the V = kQ1/r equation. |
| Remember that the radius for the center of the square is... | r = ((side length) * sqrt 2)/2 |
| If the charges are the same at each corner of the square then... | Just do 4kQ1/r for your value of total V |
| If the charges are cancel out at each corner of the square then... | V total = 0 |
| How to determine the electric potential at a specific charge on a square? | You would set up your EPE = Q1V equation and then solve for V total. Let's say you're solving for Q1 -> then V = V2 + V3 + V4. Put the V back into the equation and get the EPE. |
| If you're straight across from one corner of the square all the way to one located diagonal to it... | side value * sqrt 2 |
| Work Done by Electric Forces (FORMULAS GIVEN ON SHEET): | when a charged particle moves in a region of space where an electric field is present, the electric field exerts a force on the particle and thus does work. Work done depends only on the initial and final potential energy. Electric force is conservative |
| Units for an Electron Volt: | An electron volt (eV) is a unit of energy that is gained by an electron, which is accelerated through a potential difference of 1 volt. 1 eV = 1.6 * 10^-19 Joules. |
| How to determine the work needed by the electric field force to displace a charge from one corner to another? | Set the current position of the charge as i and where it's trying to go as f. Use the Wfe = -q(Vf - Vi). To find Vi add up the kQ1r1's of any other charges acting on it. Do the same for the Vf (remember that radiuses would change!). Put into work equation |
| Electric Potential Energy of a Pair of Point-Like Charges: | Since energy cannot be created nor destroyed, the minimum amount of work to bring these two charges near each other is going to convert into electric potential energy carried by both charges. Any more and it would convert into kinetic energy. |
| Overall Pair of Point-Like Charges Concept (FORMULA GIVEN ON SHEET) | Meke sure that the total electric potential energy of a system of charges is based on the number of pairs that form the system. Do not add up the electric potential energy of each charge because you are going to wind up with twice as much energy. |
| How to find the speed of two charges initially released from rest with identical masses and charges? | ri can be a given value and rf is infinity; Wnc = 0, so you can use the KEi + PEi formula. They'd both have separate KEs but one PE (cause same charges). Set Kei to 0 for rest and Pef to 0 for infinity. Solve for vf. |
| How to determine the potential energy of the entire system with four charges? | EPE system = kQ1Q2/r + every other possible pair for kQ/r. A square with four charges tends to have six pairs. Remember that the radius changes for diagonals! Cancel any charges with same magnitude, opposite signs. |
| Equipotential Lines | lines connecting points where electric potential is the same. The also true for equipotential surfaces → each point on the surface is at the same potential. Equipotential lines are perpendicular to the direction of the electric field lines. |
| Relationship Between Potential and Electric Field (GIVEN ON FORMULA SHEET) | E = -delta V / delta S. Delta S: the component of the displacement along the direction of the electric field force. Not the difference between the two points where we're trying to evaluate the potential difference. |
| How to calculate the magnitude and direction of the electric field between two capacitor plates and the mass of the particle in between them? | Use the E = -delta V / delta S, because delta S is the distance between the plates. delta V would be the potential difference between the plates; put both values into the original formula. Once you have E, do Fe - mg = 0, Fe = mg, solve for m. |
| Capacitors: | Consists of two conductors separated by an insulator known as a dielectric. The capacitance of a capacitor depends on the size of the plates as well as the distance between the plates. |
| Dielectric Material | to prevent the plates from touching, a dielectric material is squeezed in between them. It has its own constant called k = 1.0006 at 20 degrees celsius. The smallest value it can have is 1 for dry air, but it’s larger than 1 for all other materials. |
| Tip about Capacitance Relationships (FORMULA GIVEN IN SHEET) | Capacitance C is directly proportional to the area (A). The capacitance (C) is inversely proportional to distance (d). |
| Capacitance | The ability of a capacitor to store electric charge is referred to as capacitance (C). Q = CV (Q is the charge stored in Coulombs and V is the potential difference between the conducting surface) |
| Energy Stored in a Capacitor Formula: | A charged capacitor stores energy. The energy stored in charging a capacitor from an uncharging condition to a charge of (Q) and a potential difference (V) is given by three separate equations → ALL ARE FOUND IN THE FORMULA SHEET |
| Capacitor Demonstration | A generator charges one plate positively, which induces a neg charge on the grounded plate by attracting e-. A metal ball placed between them repeatedly moves back and forth as it alternates between pos and negative charge due to contact with the plates. |
| Discharging Capacitors Demo | Two capacitors are charged in parallel to store a large amount of energy, then rapidly discharged by closing the circuit with a metal rod across the bolts. This sudden release of stored energy produces a bright spark and loud bang. |
| How to determine the capacitance, charge of each plate, energy stored, and electric field between each plate? | For capacitance, you'd use C = Ke0A/d, K = 1 if you have dry air. For charge, use Q = CV. For energy, use any of the energy equations. For electric field between each plate, E = -delta V/delta S. S would be the distance between the plates. |
| Unit of Capacitance | 1 F = 1 C/V |
| F to pF: | Multiply by 10^12. |
| pF to F: | Divide by 10^12 or multiply by 10^-12 |
| How to determine the points on the axis where the electric potential is zero | Set up to solve for V total, use the kQ/r equations for each of the charges, put V total as 0, cross multiply to find the relationship between the r's, use the distances to write a separate equation, sub, solve |
| Tip about determining where the electric potential is zero | Remember that if the two charges are of the same kind, due to it being a scalar quantity, there is NO point along the axis where the electric potential will equal zero except for infinity. |
| How to find the electric potential at a certain point (P) on a equilateral triangle | Set up for V total, use kQ/r for all charges, remove the charges that cancel each other out, solve |
| When you see the word external force in a work question, that means... | You're dealing with a non-conservative force |
| When you see change in PE on a work question, that means... | You can equate that to q delta V |
| Whenever you start from rest... | KEi doesn't exist |
| When the question is related to work done by external forces and the work mentioned is the minimum amount of work, then... | The change in kinetic energy would be zero and all the work done would be to overcome the change in potential energy only. |
| How would you find the minimum amount of work needed by an external force to bring a charge from infinity to a certain point? | You would use the Wnc equation, remember there's ONLY potential energy, do Wfext = -q(vf-Vi). You would have to set up the Vi = kq1/r1 + kq2/q2 (both of which would equal infinity), then Vf the same way. Put back into the -q(vf-vi) equation and solve |
| The question gives you just a uniform electric field with a certain magnitude and asks you to find the electric potential difference. | You would use the E = -delta V/delta S. Delta V = Vlower - V higher. Delta V = -EdeltaS, so if the points are located straight across - no higher or lower - S = 0 and so does V. The component of S should be perfectly parallel, not hypothenuse. |
| If the spacing between the capacitor places is doubled, what is the new voltage between the plates of the capacitor? | Set up C2V2 = C1V2, then sub Kepisoln0A/2d for C2 and the regular Kepilson for C1. Solve, then notice that V2 = twice the original V battery. |
| Important Tip about Battery Connection | If modifications are made to the capacitor after the battery is disconnected, the charge will stay the same. If the modification is made to the capacitor with the battery still connected, the voltage stays the same. |
| Radius of the center of a rectangle | (side^2 )/2 + (side^2)/2 in a sqrt |
Created by:
smurtab