GenPhysics - 3rd Qtr. 2024 PDF

GENERAL PHYSICS: ELECTROSTATICS ELECTRIC CHARGE | CAPACITANCE | ELECTRICAL FLUX | ELECTRIC POTENTIAL AND POTENTIAL ENERGY SHS | STEM | SECOND SEMESTER | 2024 ELECTRIC CHARGE 𝑄...

GENERAL PHYSICS: ELECTROSTATICS ELECTRIC CHARGE | CAPACITANCE | ELECTRICAL FLUX | ELECTRIC POTENTIAL AND POTENTIAL ENERGY SHS | STEM | SECOND SEMESTER | 2024 ELECTRIC CHARGE 𝑄 𝑛=  Determines the electric interaction and 𝑒 magnetic interaction between subatomic 1𝐶 𝑛= particles and other charges. 1.6𝑥10−19 𝐶  The amount of electricity on a body. 𝑛 = 6.25𝑥1018 𝑒  A property of protons and electrons. Final Answer 𝒏 = 𝟔. 𝟐𝟓𝒙𝟏𝟎𝟏𝟖 𝒆 CHARGE EXAMPLE 2  Fundamental property of subatomic particles Aluminum has atomic number of 13. What is the  Unit: C – Coulombs (named after French total charge of all electrons in the aluminum atom? physicist Charles Agustine de Coulumb) 𝑛 = 13 𝟏 𝑪 = 𝟔. 𝟐𝟒𝟐𝒙𝟏𝟎𝟏𝟖 𝒆 Given 𝑒 = 1.6𝑥10−19 𝐶 ELEMENTARY CHARGE, e Asked 𝑄 =?  Smallest value of charge. Formula 𝑄 = 𝑛𝑒 𝑄 = 13(−1.6𝑥10−19 𝐶) 𝒆 = 𝟏. 𝟔𝟎𝒙𝟏𝟎−𝟏𝟗 𝑪 Solution 𝑄 = −2.08𝑥10−18 𝐶 3 PROPERTIES OF CHARGE Final Answer 𝑸 = −𝟐. 𝟎𝟖𝒙𝟏𝟎−𝟏𝟖𝑪 1. Charges of the same sign repel on another, and charges of the opposite sign attract one 3. Charge is Conserved another  Charges are neither created nor destroyed.  There is the same amount of charge in the universe now as there has always been. Conductors  Materials in which some of the electrons are free  Two charges with electrons that are NOT the opposite signs bound to atoms and can  A positive charge attract move freely through the repels another material positive charge  Examples: copper,  A negative charge aluminum, silver repels another Insulators  Materials in which all of negative charge the electrons are bound to atoms and CANNOT 2. Charge is Quantized move freely through the  Charge is zero or a multiple of elementary materials. charge, e.  Examples: glass, wood, 𝑸 = 𝒏𝒆 rubber  𝑄 – variable charge in Coulombs, C  𝑛 – number of electrons/protons CHARGING  𝑒 – elementary charge, 1.6𝑥10−19 𝐶  The process of transferring electrons from one object to another. EXAMPLE 1 Friction  Rubbing things together, How many electrons must an object lose so that it electrons can peel off one has a net positive charge of 1 C? material and remains in the other. 𝑄=1𝐶 Conduction  A charged object comes Given 𝑒 = 1.6𝑥10−19 𝐶 into contact with another Asked 𝑛 =? object, electrons are Formula 𝑄 = 𝑛𝑒 transferred, thereby 𝑄 = 𝑛𝑒 charging the second Solution object. Derived: fau GENERAL PHYSICS: ELECTROSTATICS ELECTRIC CHARGE | CAPACITANCE | ELECTRICAL FLUX | ELECTRIC POTENTIAL AND POTENTIAL ENERGY SHS | STEM | SECOND SEMESTER | 2024 Induction  There is no actual contact  Suppose a strongly negative body B is between the charged brought near A. The negative charges of object and that which is A will move to the side farther from where being charged. B is. Consequently, the positive charges of A will be attracted toward the side near 1. CHARGING BY RUBBING/FRICTION B. If B will touch A, some of B’s negative  An electrically neutral body can gain a charge charges will transfer to A as these spread by rubbing or friction. Consider two different away from each other. If B will be uncharged bodies. Because of the difference separated from A, B is now less negative in their material compositions, the nuclei of than before, while A is now more negative. their atoms pull their electrons with different strengths.  Rubbing these two bodies will force their atoms to interact with one another, resulting in the “ripping off” of electron(s) from the body with a weaker electron hold.  The “ripped off” electrons are then transferred to the other body. 3. CHARGING BY INDUCTION  After rubbing, one of the bodies will have  A negatively charged balloon is placed near more electrons, and the other one will have (but without physical contact) a neutral tin fewer. Thus, both of them will now be can. As a result, the positive charges of the electrically charged can move to the left, near the negatively charged balloon because of attraction.  Meanwhile, the negative charges move to the right side of the tin can, away from the balloon because of repulsion. The negative charges at the right side of the tin can are known as induced charges. TRIBOELECTRIC CHARGING  If you rub two materials from the list against each other, the one nearer the top will receive net positive charge and the other a net negative charge.  More intense rubbing creates greater charge GROUNDING transfer.  Grounding is a process like conduction, but it includes a grounding wire that connects the neutral body A to the ground, which is a 2. CHARGING BY CONDUCTION reservoir of charge.  A body can also be electrically charged  If a strongly positive body B is brought near A, through conduction. Consider a neutral body the negative charges of A will move to the side A. Charges in A are evenly distributed near B. Negative charges from the ground will throughout the material. also travel to the same side via the grounding fau GENERAL PHYSICS: ELECTROSTATICS ELECTRIC CHARGE | CAPACITANCE | ELECTRICAL FLUX | ELECTRIC POTENTIAL AND POTENTIAL ENERGY SHS | STEM | SECOND SEMESTER | 2024 wire. If the grounding wire is suddenly cut, the Inverse Square Rule negative charges that went to A from the  If the distance is multiplied by 2, the force ground will stay in A, thus making the body will be divided by 4 more negatively charged. 2𝑟 2 = 22 = 𝟒  If the distance is multiplied by 3, the force will be divided by 9 3𝑟 2 = 32 = 𝟗 Determining the Direction of the Electrical Force Vector ELECTRIC DIPOLES  An electric dipole is a pair of equal and opposite electric charges separated by a small distance.  This separation creates a moment called the dipole moment, representing the strength and orientation of the dipole. Electric dipoles experience a phenomenon known as polarization when subjected to an external electric field.  In response to the field, the positive and Naming Electrostatic Force negative charges within the dipole shift, causing a temporary alignment that enhances the overall electric moment.  This induced polarization is a crucial concept in understanding the behavior of dielectric materials and has applications in various technologies, including capacitors and molecular interactions in physics and chemistry. EXAMPLE 1 (Coulomb’s Law) Two charged spheres 𝑄1 = −2.00 𝑥 10−10 𝐶 and 𝑄2 = −5.00 𝑥 10−8 𝐶 held fixed at positions 5.00 𝑐𝑚 apart. Calculate the magnitude of each electrostatic force between the two spheres. Are COULOMB’S LAW the forces attractive or repulsive?  “The force in Newton's between two charges is directly proportional to the product of the 𝑄1 = −2𝑥10−10 𝐶 magnitudes of the charges and inversely 𝑄2 = −5𝑥10−8 𝐶 Given proportional to the square of the distance 𝑘 = 9𝑥109 𝑁 ∙ 𝑚2 /𝐶 2 between them.” 𝑟 = 5.00 𝑐𝑚 = 0.05 𝑚 |𝑸 𝟏 𝑸 𝟐 | Asked 𝐹𝑒 → 𝐹21 , 𝐹12 𝑭𝒆 = 𝒌 𝒓𝟐 Formula |𝑄 𝑄 | 𝐹𝑒 = 𝑘 1 2 2  𝑘 – Coulomb’s constant (electrostatic 𝑟 constant) = 9𝑥109 𝑁 ∙ 𝑚2 /𝐶 2 𝐹𝑒 (9𝑥109 )|(−2𝑥10−10 )(−5𝑥10−8 )|  𝑄1 𝑄2 – charges in Coulomb, C Solution = (0.05)2  𝑟 – distance between the charges in 𝐹𝑒 = 3.6𝑥10−5 𝑁 meters, m Final Answer 𝑭𝒆 = 𝟑. 𝟔𝒙𝟏𝟎−𝟓𝑵  𝐹𝑒 – electrostatic force in Newtons, N fau GENERAL PHYSICS: ELECTROSTATICS ELECTRIC CHARGE | CAPACITANCE | ELECTRICAL FLUX | ELECTRIC POTENTIAL AND POTENTIAL ENERGY SHS | STEM | SECOND SEMESTER | 2024 EXAMPLE 2 (Coulomb’s Law)  An electric field coexists with every What is the electrostatic force of attraction between electrostatic charge; it associates with each a −6.0𝑥10−6 𝐶 charge and a 4.0𝑥10−6 𝐶 charge if point in space the electrostatic force they are separated by a distance of 3 𝑚𝑒𝑡𝑒𝑟𝑠 (𝑚)? experienced per unit of electric charge, by an extremely small (or infinitesimal) test charge 𝑄1 = −6.0𝑥10−6 𝐶 at that point. 𝑄2 = 4.0𝑥10−6 𝐶 Given 𝑘 = 9𝑥109 𝑁 ∙ 𝑚2 /𝐶 2 𝑟=3𝑚 Asked 𝐹𝑒 = ? |𝑄 𝑄 | Formula 𝐹𝑒 = 𝑘 1 2 2 𝑟 𝐹𝑒 (9𝑥109 )|(−6.0𝑥10−6 )(4.0𝑥10−6 𝐶)| Solution = (3)2 𝐹𝑒 ≈ 0.02 𝑁 Final Answer 𝑭𝒆 ≈ 𝟎. 𝟎𝟐 𝑵 MAGNITUDE OF ELECTRIC FIELD EXAMPLE 3 (Coulomb’s Law) 𝑭𝒆 |𝑸 | Two identically charged one-peso coins are 1.5 𝑚 𝑬= ; 𝑬=𝒌 𝟐 𝒒 𝒓 apart on a table. What is the charge of the coins if each of them experiences a repulsive force of 2.0 𝑁?  𝐸 – electric field  𝐹𝑒 – electrostatic force, Newtons (N) 𝑘 = 9𝑥109 𝑁 ∙ 𝑚2 /𝐶 2  𝑞, 𝑄 – charge, Coulombs (C) Given 𝑟 = 1.5 𝑚  𝑟 – distance, meters (m) 𝐹𝑒 = 2.0 𝑁  𝑘 – Coulomb's constant (electrostatic constant) = 9 x 109 N·m2/C2 Asked 𝑞=? |𝑄 𝑄 | Formula 𝐹𝑒 = 𝑘 1 2 2  Unit: Newtons/Coulomb, N/C 𝑟 2  Note: |𝑄1 𝑄2 | |𝑄 | 𝐹𝑒 = 𝑘 = 𝐹𝑒 = 𝑘 2 For a positive test charge, E and Fe have the 𝑟2 𝑟 Derive: same direction. Solution For a negative test charge, E and Fe are in (2.0 𝑁)(1.5 𝑚2 ) opposite direction. 𝑞=√ 9𝑥109 𝑁 ∙ 𝑚2 /𝐶 2 𝑞 ≈ 2.23𝑥10−5 𝐶 EXAMPLE 1 Final Answer 𝒒 ≈ 𝟐. 𝟐𝟑𝒙𝟏𝟎−𝟓 𝑪 Calculate the electric field that a test charge will experience on the following distances from the ELECTRIC FIELD source charge of +5.02𝑥10−13 𝐶 a. Distance from source charge: 2.04𝑥10−3 𝑚  A region around a charge where electric force acts. 𝑘 = 9𝑥109 𝑁 ∙ 𝑚2 /𝐶 2  Represented by a series of lines called electric field lines. Given 𝑟 = 2.04𝑥10−3 𝑚  The lines (arrows) leave a positive charge and 𝑄 = 5.02𝑥10−13 𝐶 enter a negative charge. Asked 𝐸 =?  The spacing determines the strength of the |𝑄| Formula 𝐸=𝑘 2 field. 𝑟 𝑁𝑚2  Field is strong if lines are close together, weak 9𝑥109 (5.02𝑥10−13 𝐶) Solution 𝐸= 𝐶2 if lines are far apart. 2 (2.04𝑥10−3 𝑚)  The area or field around a charge where the The source charge will electrostatic force can be experienced is Final Answer experience an electric field of called the electric field. 𝑬 ≈ 𝟏𝟎𝟖𝟒. 𝟒𝟑 𝑵/𝑪 fau GENERAL PHYSICS: ELECTROSTATICS ELECTRIC CHARGE | CAPACITANCE | ELECTRICAL FLUX | ELECTRIC POTENTIAL AND POTENTIAL ENERGY SHS | STEM | SECOND SEMESTER | 2024 EXAMPLE 2 EXAMPLE 3 A charge of +3.0𝑥10−8 𝐶 experiences an The charge Q has a magnitude of 2.00 𝑥 10−6 𝐶 electrostatic force of 6.0𝑥10−8 𝑁. Compute the force a. Calculate the strength of the electric field at per coulomb that the charge experiences. point A which is 0.700 𝑚 from Q b. What is the direction of the electric field at 𝐹𝑒 = 6.0𝑥10−8 𝑁 point A? Given c. Calculate the magnitude of the electrostatic 𝑄 = 3.0𝑥10−8 𝐶 Asked 𝐸 =? force exerted by Q on a charge 𝑞 = 𝐹𝑒 2.00 𝑥 10−20 𝐶 placed at A Formula 𝐸= d. What is the direction of this force? 𝑞 6.0𝑥10−8 𝑁 Solution 𝐸= 3.0𝑥10−8 𝐶 𝐸 = 2.0 𝑁/𝐶 The force per coulomb of the charge or the associated Final Answer electric field is 𝑄 = 2.0𝑥10−6 𝐶 𝑬 = 𝟐. 𝟎 𝑵/𝑪 Given 𝑟 = 0.700 𝑚 𝑘 = 9𝑥109 𝑁𝑚2 /𝐶 2 EXAMPLE 2 Asked a. 𝐸 = ? Compute the electric field experienced by a test b. 𝐷𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝐸 = ? charge 𝑞 = +0.80 𝜇𝐶 from a source charge 𝑞 = |𝑄| +15 𝜇𝐶 in a vacuum when the test charge is placed Formula 𝐸=𝑘 2 𝑟 0.20 𝑚 away from the other charge. 𝑁𝑚2 9𝑥109 (2.0𝑥10−6 𝐶) 𝐸= 𝐶2 𝑞 = 0.80 𝜇𝐶 = 0.8𝑥10−6 𝐶 Solution (0.700 𝑚)2 𝑄 = 15 𝜇𝐶 = 15𝑥10−6 𝐶 𝐸 = 36,734 𝑁/𝐶 Given 𝑘 = 9𝑥109 𝑁 ∙ 𝑚2 /𝐶 2 𝑬 = 𝟑𝟔, 𝟕𝟑𝟒 𝑵/𝑪 𝑟 = 0.20 𝑚 Final Answer 𝑫𝒊𝒓𝒆𝒄𝒕𝒊𝒐𝒏 𝒐𝒇 𝑬: 𝑻𝒐𝒘𝒂𝒓𝒅𝒔 𝒑𝒕 𝑨 Asked 𝐹𝑒 = ? /𝒍𝒆𝒇𝒕 |𝑄 𝑄 | 𝐹𝑒 = 𝑘 1 2 2 Formula 𝑟 𝐹𝑒 𝐸= 𝑞 = 2.0𝑥10−20 𝐶 𝑞 Given 𝐸 = 36,734 𝑁/𝐶 𝐹𝑒 (9𝑥109 )|(0.8𝑥10−6 )(15𝑥10−6 𝐶)| Asked c. 𝐹𝑒 = ? = d. 𝐷𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝐹𝑒 = ? (0.20)2 𝐹𝑒 = 2.7 𝑁 𝐹𝑒 Solution Formula 𝐸= 𝑞 2.7 𝑁 𝐹𝑒 = 𝐸𝑞 𝐸= Solution 𝐹𝑒 = (36,734 𝑁/𝐶)(2.0𝑥10−20 𝐶) 0.8𝑥10−6 𝐶 𝐸 = 3.4𝑥10−6 𝑁/𝐶 𝐹𝑒 = 7.35𝑥10−16 𝑁 Final Answer 𝑬 = 𝟑. 𝟒𝒙𝟏𝟎−𝟔 𝑵/𝑪 𝑭𝒆 = 𝟕. 𝟑𝟓𝒙𝟏𝟎−𝟏𝟔 𝑵 Final Answer 𝑫𝒊𝒓𝒆𝒄𝒕𝒊𝒐𝒏 𝒐𝒇 𝑭𝒆 : 𝑺𝒂𝒎𝒆 𝒂𝒔 𝑬 /𝒍𝒆𝒇𝒕 fau GENERAL PHYSICS: ELECTROSTATICS ELECTRIC CHARGE | CAPACITANCE | ELECTRICAL FLUX | ELECTRIC POTENTIAL AND POTENTIAL ENERGY SHS | STEM | SECOND SEMESTER | 2024 CAPACITANCE CAPACITANCE FORMULA  Is measure of the ability of the capacitor to  The amount of charge a capacitor can store store electrical energy. per unit voltage is quantified by its  Is a fundamental concept in physics that plays capacitance. a crucial role in the study of electrical circuits.  Mathematically, capacitance (C) is defined as  It is the ability of a system to store electrical the ratio of the stored charge (Q) to the charge when a voltage difference exists applied voltage (V). between its conductive surfaces. 𝑸  The basic unit of capacitance is the farad (F), 𝑪= 𝑽 named after the pioneering physicist Michael Faraday. Unit: 𝒄𝒐𝒖𝒍𝒐𝒎𝒃 = 𝑭𝒂𝒓𝒂𝒅 𝒗𝒐𝒍𝒕 CAPACITOR  A device used to store electrical charge.  𝐶 – capacitance, Farad (F)  Only for temporary storage; cannot be  𝑄 – charge on either plate, Coulomb (C) used as the main power supply.  𝑉 – electric potential (potential difference),  One of its main characteristics is that it Volts (V) easily charges, indicating that it also easily discharges. Power of 10 Prefix Abbreviation  Provides a boost of energy in a portion of 1012 tera 𝑇 109 giga 𝐺 the circuit. 106 mega 𝑀  Consists of two conductive plates separated 103 kilo 𝑘 by a dielectric material. 10−3 milli 𝑚  The dielectric material is an insulator. 10−6 micro 𝜇  When a voltage is applied across the plates, 10−9 nano 𝑛 electrons accumulate on one plate, creating a 10−12 pico 𝑝 negative charge, while the other plate  The typical values of capacitance are in micro experiences a positive charge. Farads and pico Farads.  The bigger the capacitor, the higher the capacitance CAPACITANCE BASED ON THE PHYSICAL DIMENSIONS 𝑨𝜺𝟎 𝑪= 𝒅  𝐴 – area of the conductor, 𝑚2  𝜀0 – permittivity (electric constant) = 8.85𝑥10−12 𝐹/𝑚  𝑑 – distance between the two conductors  Ideally, the construction of a capacitor should involve conductor plates that are as wide as possible, and the gap or distance between DIELECTRIC these two conductors should as small as  A capacitor is a circuit component that possible. temporarily stores charges within the circuit.  Inside it are two conducting plates facing each FACTORS AFFECTING CAPACITANCE other and separated by an insulator referred Variable Effect to as a dielectric. Area of the Increase Increase conducting plate Decrease Decrease  This material impedes the continuous Distance between Increase Decrease passage of electric current through the conducting plates Decrease Increase capacitor and stores it until it is discharges More conducting Decrease later. Type of dielectric Less conducting Increase  More dielectric, more/higher capacitance. fau GENERAL PHYSICS: ELECTROSTATICS ELECTRIC CHARGE | CAPACITANCE | ELECTRICAL FLUX | ELECTRIC POTENTIAL AND POTENTIAL ENERGY SHS | STEM | SECOND SEMESTER | 2024 EXAMPLE 1 SHAPES OF CAPACITOR Compute the capacitance if the value of charge is  Capacitors can be classified in terms of their 0.3𝑥10−6 𝐶 and the voltage supplied is 1𝑥103 𝑉. construction–parallel-plate, spherical, and cylindrical. Each of there has advantages and 𝑄 = 0.3𝑥10−6 𝐶 disadvantages based on capacitance, Given 𝑉 = 1𝑥103 𝑉 charge, and potential difference. Asked 𝐶 =? 𝑄 1. PARALLEL-PLATE CAPACITOR Formula 𝐶= 𝑉  In this type of capacitor, two parallel charging 0.3𝑥10−6 𝐶 plates are separated by a dielectric that 𝐶= Solution 1𝑥103 𝑉 contains the charges. 𝐶 = 3𝑥10−10 𝐹  The capacitance that can be offered by a parallel-plate capacitor is directly proportional Final Answer 𝑪 = 𝟑𝒙𝟏𝟎−𝟏𝟎 𝑭 to the area of the plates as well as to the distance between these plates. The voltage EXAMPLE 2 across this type of capacitor is also directly Calculate the capacitance of a capacitor if it stores proportional to the distance between the a charge of 4.1𝑥10−5 𝐶 when delivered with a plates. voltage of 2.5𝑥104 𝑉. 𝑄 = 4.1𝑥10−5 𝐶 Given 𝑉 = 2.5𝑥104 𝑉 Asked 𝐶 =? 𝑄 2. CYLINDRICAL CAPACITOR Formula 𝐶= 𝑉  Cylindrical capacitors have a different 4.1𝑥10−5 𝐶 𝐶= construction compared to a parallel-plate. As Solution 2.5𝑥104 𝑉 seen in the figure, inner and outer cylindrical 𝐶 = 1.64𝑥10−9 𝐹 structures correspond to the plates of the Final Answer 𝑪 = 𝟏. 𝟔𝟒𝒙𝟏𝟎−𝟗 𝑭 parallel-plate capacitor. The dielectric is placed between these two charged cylinders. EXAMPLE 3 A battery connected to the plates of a 3.00 𝜇𝐹 capacitor stores a charge of 27.0 𝜇𝐶. How much voltage is provided by the battery? 𝑄 = 27.0 𝜇𝐶 = 27𝑥10−6 𝐶 Given 𝐶 = 3.00 𝜇𝐹 = 3𝑥10−6 𝐹 Asked 𝑉=? 3. SPHERICAL CAPACITOR 𝑄 Formula 𝐶=  Spherical capacitors have a construction like 𝑉 Derive: that of a cylindrical capacitor. As seen in the 𝑄 figure, an internal spherical structure is one of 𝑉= the charged bodies of the capacitor. The other 𝐶 Solution 27𝑥10−6 𝐶 charged body is the outer spherical structure 𝑉= that covers the internal sphere. The dielectric 3𝑥10−6 𝐹 is places between these two charged 𝑉 =9𝑉 spheres. Final Answer 𝑽=𝟗𝑽 fau GENERAL PHYSICS: ELECTROSTATICS ELECTRIC CHARGE | CAPACITANCE | ELECTRICAL FLUX | ELECTRIC POTENTIAL AND POTENTIAL ENERGY SHS | STEM | SECOND SEMESTER | 2024 CAPACITOR IN A CIRCUIT  Capacitor’s function when they are connected to a circuit. Separate treatments should be done for capacitors connected in series and in parallel. CAPACITORS IN SERIES  The main characteristics of a series circuit are: SUMMARY 1. Same current flows through all parts of SERIES the circuit. In capacitors in series the following rules apply: 2. Different components have their  𝑉𝑇 = 𝑉1 + 𝑉2 + 𝑉3 … + 𝑉𝑁 individual voltage drops.  𝑄𝑇 = 𝑄1 = 𝑄2 = 𝑄3 … = 𝑄𝑁 3. Voltage drops are additive. 1 1 1 1 1  = + + …+ 𝐶𝑒𝑞 𝐶1 𝐶2 𝐶3 𝐶𝑁 4. Applied voltage equals the sum of different voltage drops.  Capacitors in series: PARALLEL  The capacitors have the same charge 𝑄 In capacitors in parallel the following rules apply:  Their potential differences add:  𝑉𝑇 = 𝑉1 = 𝑉2 = 𝑉3 … = 𝑉𝑁 𝑉𝑎𝑐 + 𝑉𝑐𝑏 = 𝑉𝑎𝑏  𝑄𝑇 = 𝑄1 + 𝑄2 + 𝑄3 … + 𝑄𝑁  𝐶𝑒𝑞 = 𝐶1 + 𝐶2 + 𝐶3 … + 𝐶𝑁 EXAMPLE 1 (Capacitors in Series) Compute the individual values and the total values of the capacitance, charge, and voltage of the circuit in the figure. 𝐶1 = 2.5𝑥10−12 𝐹 CAPACITORS IN PARALLEL Given 𝐶2 = 4.1𝑥10−12 𝐹  The main characteristics of a parallel circuit 𝑉𝑇 = 1500 𝑉 are: Asked 𝑄1 , 𝑄2 , 𝑄𝑇 , 𝑉1 , 𝑉2 , 𝐶𝑒𝑞/𝑇 1. Same voltage acts across all parts of the 𝑄 𝑄 circuit. Formula 𝐶 = ; 𝑉 = ; 𝑄 = 𝐶𝑉 𝑉 𝐶 2. Different components have their 1 individual current. 𝐶𝑇 = 1 1 3. Branch currents are additive. + 𝐶1 𝐶2  Capacitors in parallel: 1  The capacitors have the same potential 𝑉 𝐶𝑇 = 1 1  The charge on each capacitor depends on + 2.5𝑥10−12 4.1𝑥10−12 its capacitance: 𝑄1 = 𝐶1 𝑉, 𝑄2 = 𝐶2 𝑉 Solution 𝑪𝑻 = 𝟏. 𝟓𝟓𝒙𝟏𝟎−𝟏𝟐 𝑭 𝑄𝑇 = 𝐶𝑇 𝑉𝑇 𝑄𝑇 = (1.55𝑥10−12)(1500) 𝑸𝑻 = 𝟐. 𝟑𝟑𝒙𝟏𝟎−𝟗 𝑪 = 𝑸𝟏 = 𝑸𝟐 fau GENERAL PHYSICS: ELECTROSTATICS ELECTRIC CHARGE | CAPACITANCE | ELECTRICAL FLUX | ELECTRIC POTENTIAL AND POTENTIAL ENERGY SHS | STEM | SECOND SEMESTER | 2024 2.33𝑥10−9 𝐶 𝑄𝑇 = 𝐶𝑇 𝑉𝑇 𝑉1 = 2.5𝑥10−12 𝐹 𝑄𝑇 = (8.2𝑥10−12 𝐹)(1600 𝑉) 𝑽𝟏 = 𝟗𝟑𝟐 𝑽 𝑸𝑻 = 𝟏. 𝟑𝟏𝒙𝟏𝟎−𝟖 𝑪 𝑄2 𝑸𝟏 = 𝟓. 𝟔𝒙𝟏𝟎−𝟗 𝑪 𝑉2 = 𝐶2 𝑸𝟐 = 𝟕. 𝟓𝟐𝒙𝟏𝟎−𝟗 𝑪 2.33𝑥10−9 𝐶 𝑸𝑻 = 𝟏. 𝟑𝟏𝒙𝟏𝟎−𝟖 𝑪 𝑉2 = Final Answer 4.1𝑥10−12 𝐹 𝑽𝟏 = 𝟏𝟔𝟎𝟎 𝑽 𝑽𝟐 = 𝟓𝟔𝟖. 𝟐𝟗 𝑽 𝑽𝟐 = 𝟏𝟔𝟎𝟎 𝑽 𝑪𝑻 = 𝟖. 𝟐𝒙𝟏𝟎−𝟏𝟐 𝑭 𝑸𝟏 = 𝟐. 𝟑𝟑𝒙𝟏𝟎−𝟗 𝑪 𝑸𝟐 = 𝟐. 𝟑𝟑𝒙𝟏𝟎−𝟗 𝑪 EXAMPLE 3 (Combination) 𝑸𝑻 = 𝟐. 𝟑𝟑𝒙𝟏𝟎−𝟗 𝑪 Final Answer 𝑽𝟏 = 𝟗𝟑𝟐 𝑽 𝑽𝟐 = 𝟓𝟔𝟖. 𝟐𝟗 𝑽 12 V 𝑪𝑻 = 𝟏. 𝟓𝟓𝒙𝟏𝟎−𝟏𝟐 𝑭 EXAMPLE 2 (Capacitors in Parallel) Compute the individual values and the total values of the capacitance, charge, and voltage of the given circuit. 𝐶1 = 3.5𝑥10−12 𝐹 Given 𝐶2 = 4.7𝑥10−12 𝐹 𝑉𝑇 = 1600 𝑉 Asked 𝑄1 , 𝑄2 , 𝑄𝑇 , 𝑉1 , 𝑉2 , 𝐶𝑒𝑞 𝑄 𝑄 Formula 𝐶 = ; 𝑉 = ; 𝑄 = 𝐶𝑉 𝑉 𝐶 𝐶𝑇 = 𝐶1 + 𝐶2 𝐶𝑇 = 3.5𝑥10−12 + 4.7𝑥10−12 𝑪𝑻 = 𝟖. 𝟐𝒙𝟏𝟎−𝟏𝟐 𝑭 𝐶1 = 4𝜇𝐹 = 4𝑥10−6 𝐹 𝐶2 = 2𝜇𝐹 = 2𝑥10−6 𝐹 𝐶3 = 3𝜇𝐹 = 3𝑥10−6 𝐹 𝑉𝑇 = 𝑉1 = 𝑉2 Given 𝐶4 = 1𝜇𝐹 = 1𝑥10−6 𝐹 𝑽𝑻 = 𝟏𝟔𝟎𝟎 𝑽 = 𝑽𝟏 = 𝑽𝟐 𝐶5 = 6𝜇𝐹 = 6𝑥10−6 𝐹 Solution 𝑄1 = 𝐶1𝑉1 𝑉𝑇 = 12 𝑉 Asked 𝐶𝑇 , 𝑉1 , 𝑉2 , 𝑉3 , 𝑉4 , 𝑉5 , 𝑄1 , 𝑄2 , 𝑄3 , 𝑄4 , 𝑄5 , 𝑄𝑇 𝑄1 = (3.5𝑥10−12 𝐹)(1600 𝑉) 𝑸𝟏 = 𝟓. 𝟔𝒙𝟏𝟎−𝟗 𝑪 𝑄 𝑄 Formula 𝐶= ; 𝑉 = ; 𝑄 = 𝐶𝑉 𝑉 𝐶 1 𝑄2 = 𝐶2 𝑉2 𝐶34 = 1 1 𝑄2 = (4.7𝑥10−12 𝐹)(1600 𝑉) 𝐶3 + 𝐶4 𝑸𝟐 = 𝟕. 𝟓𝟐𝒙𝟏𝟎−𝟗 𝑪 Solution 1 𝐶34 = 1 1 + 3𝑥10−6 1𝑥10−6 𝑪𝟑𝟒 = 𝟕. 𝟓𝒙𝟏𝟎−𝟕𝑭 fau GENERAL PHYSICS: ELECTROSTATICS ELECTRIC CHARGE | CAPACITANCE | ELECTRICAL FLUX | ELECTRIC POTENTIAL AND POTENTIAL ENERGY SHS | STEM | SECOND SEMESTER | 2024 1 9𝑥10−6 𝐶 𝐶12 = 𝑉4 = 1 1 1𝑥10−6 𝐹 + 𝐶1 𝐶2 𝑽𝟒 = 𝟗 𝑽 1 𝐶12 = 1 1 + 𝑪𝑻 = 𝟖. 𝟎𝟖𝒙𝟏𝟎−𝟔 𝑭 4𝑥10−6 2𝑥10−6 𝑪𝟏𝟐 = 𝟏. 𝟑𝟑𝒙𝟏𝟎−𝟔 𝑭 𝑽𝟏 = 𝟒 𝑽 𝑽𝟐 = 𝟖 𝑽 𝐶𝑇 = 𝐶12 + 𝐶34 + 𝐶5 𝑽𝟑 = 𝟑 𝑽 1.33𝑥10−6 + 7.5𝑥10−7 + 6𝑥10−6 𝑽𝟒 = 𝟗 𝑽 𝑪𝑻 = 𝟖. 𝟎𝟖𝒙𝟏𝟎−𝟔 𝑭 𝑽𝟓 = 𝟏𝟐 𝑽 Final Answer 𝑽𝑻 = 𝟏𝟐 𝑽 = 𝑽𝟏𝟐 = 𝑽𝟑𝟒 = 𝑽𝟓 𝑸𝟏 = 𝟏. 𝟔𝒙𝟏𝟎−𝟓 𝑪 𝑸𝟐 = 𝟏. 𝟔𝒙𝟏𝟎−𝟓 𝑪 𝑄12 = 𝐶12 𝑉12 𝑸𝟑 = 𝟗𝒙𝟏𝟎−𝟔 𝑪 −6 𝑄12 = (1.33𝑥10 𝐹)(12 𝑉) 𝑸𝟒 = 𝟗𝒙𝟏𝟎−𝟔 𝑪 −𝟓 𝑸𝟏𝟐 = 𝟏. 𝟔𝒙𝟏𝟎 𝑪 𝑸𝟓 = 𝟕. 𝟐𝒙𝟏𝟎−𝟓 𝑪 𝑸𝑻 = 𝟗. 𝟕𝒙𝟏𝟎−𝟓 𝑪 𝑄34 = 𝐶34 𝑉34 𝑄34 = (7.5𝑥10−7𝐹)(12 𝑉) 𝑸𝟑𝟒 = 𝟗𝒙𝟏𝟎−𝟔 𝑪 𝑄5 = 𝐶5 𝑉5 ELECTRIC FLUX 𝑄5 = (6𝑥10−6 𝐹)(12 𝑉)  The amount of electric field passing through a 𝑸𝟓 = 𝟕. 𝟐𝒙𝟏𝟎−𝟓 𝑪 surface. 𝑄𝑇 = 𝐶𝑇 𝑉𝑇 𝝓 = 𝑬𝑨𝒄𝒐𝒔𝜽 𝑄𝑇 = (8.08𝑥10−6 𝐹)(12 𝑉)  𝐸 – electric field in N/C (or V/m) 𝑸𝑻 = 𝟗. 𝟕𝒙𝟏𝟎−𝟓 𝑪  𝐴 – surface area in 𝑚2 or  𝜃 – angle between the area vector, ⃗⃗⃗𝐴 and 𝑄𝑇 = 𝑄12 + 𝑄34 + 𝑄5 the electric field 𝐸 1.6𝑥10−5 + 9𝑥10−6 + 7.2𝑥10−5  Unit: 𝑁𝑚2 /𝐶 or volt-m 𝑸𝑻 = 𝟗. 𝟕𝒙𝟏𝟎−𝟓 𝑪  Area vector, ⃗⃗⃗ 𝐴 – similar to the normal force, 𝑄12 = 𝑄1 = 𝑄2 perpendicular and away from the surface 𝑸𝟏𝟐 = 𝟏. 𝟔𝒙𝟏𝟎−𝟓 𝑪 = 𝑸𝟏 = 𝑸𝟐 𝑄34 = 𝑄3 = 𝑄4 𝑸𝟑𝟒 = 𝟗𝒙𝟏𝟎−𝟔 𝑪 = 𝑸𝟑 = 𝑸𝟒 For 𝑉1 , 𝑉2 , 𝑉3 , 𝑉4 𝑄1 𝑉1 = 𝐶1 1.6𝑥10−5 𝐶 𝑉1 = 4𝑥10−6 𝐹 𝑽𝟏 = 𝟒 𝑽 a. Surface is Face-On to Electric Field 𝑄2 𝑉2 =  ⃗⃗⃗ 𝐸 and ⃗⃗⃗ 𝐴 are parallel 𝐶2 1.6𝑥10−5 𝐶  The flux 𝜙𝐸 = ⃗⃗⃗ ⃗⃗⃗ = 𝐸𝐴 𝐸 ∙𝐴 𝑉2 = 2𝑥10−6 𝐹 𝑽𝟐 = 𝟖 𝑽 𝑄3  𝜙𝐸 = 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑉3 = 𝐶3  𝜃 = 0° 𝑜𝑟 180° 9𝑥10−6 𝐶 𝑉3 = 3𝑥10−6 𝐹 𝑽𝟑 = 𝟑 𝑽 𝑄4 𝑉4 = 𝐶4 fau GENERAL PHYSICS: ELECTROSTATICS ELECTRIC CHARGE | CAPACITANCE | ELECTRICAL FLUX | ELECTRIC POTENTIAL AND POTENTIAL ENERGY SHS | STEM | SECOND SEMESTER | 2024 b. Surface is Tilted from a Face-On EXAMPLE 3 Orientation by an Angle 𝝓 A cube with sides measuring 5 cm is placed in an  The angle between 𝐸 ⃗⃗⃗ and 𝐴 ⃗⃗⃗ is 𝜙 electric field with magnitude of 3.00 x 103 N/C. The ⃗⃗⃗ ∙ 𝐴 ⃗⃗⃗ = 𝐸𝐴𝑐𝑜𝑠𝜙 electric field lines are directed horizontally from left  The flux 𝜙𝐸 = 𝐸 to right. Calculate the flux on each of the six faces of the cube. What is the total flux?  𝜙𝐸 = 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 ⃗⃗⃗ = 3.00𝑥103 𝑁/𝐶 𝐸 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑠 = 5 𝑐𝑚 = 0.05 𝑚  𝜃 = 𝑔𝑖𝑣𝑒𝑛 𝑎𝑛𝑔𝑙𝑒 Given b. Surface is Edge-On to Electric Field Asked 𝜙 𝑜𝑛 𝑒𝑎𝑐ℎ 𝑓𝑎𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑢𝑏𝑒  ⃗⃗⃗ and 𝐴 𝐸 ⃗⃗⃗ are perpendicular Formula 𝜙𝐸 = (⃗⃗⃗ 𝐸 )(𝐴)(𝑐𝑜𝑠𝜃)  The flux 𝜙𝐸 = 𝐸 ⃗⃗⃗ ∙ 𝐴 ⃗⃗⃗ = 𝐸𝐴𝑐𝑜𝑠90° = 0 Only solve for the side/faces of the cube that have electric flux:  Left and right side  𝜙𝐸 = 0 𝜙𝑟𝑖𝑔ℎ𝑡  𝜃 = 90° = (3𝑥103 𝑁/𝐶)(0.05𝑥0.05)𝑚 2𝑐𝑜𝑠0° 𝝓𝒓𝒊𝒈𝒉𝒕 = 𝟕. 𝟓 𝑵𝒎𝟐/𝑪 EXAMPLE 1 𝜙𝑙𝑒𝑓𝑡 What is the electric flux for the following sets of = (3𝑥103 𝑁/𝐶)(0.05𝑥0.05)𝑚2𝑐𝑜𝑠180° Solution variables? 𝝓𝒍𝒆𝒇𝒕 = −𝟕. 𝟓 𝑵𝒎𝟐 /𝑪 ⃗⃗⃗ = 3.4𝑥1023 𝑁/𝐶 𝐸 𝜙𝑡𝑜𝑡𝑎𝑙 Given 𝐴 = 2.3𝑥10−2 𝑚2 = 𝜙𝑡𝑜𝑝 + 𝜙𝑏𝑜𝑡𝑡𝑜𝑚 + 𝜙𝑓𝑟𝑜𝑛𝑡 𝜃 = 84° +𝜙𝑏𝑎𝑐𝑘 + 𝜙𝑟𝑖𝑔ℎ𝑡 + 𝜙𝑙𝑒𝑓𝑡 Asked 𝜙𝐸 = ? Formula 𝜙𝐸 = (⃗⃗⃗ 𝐸 )(𝐴)(𝑐𝑜𝑠𝜃) = 0 + 0 + 0 + 0 + 7.5 − 7.5 𝝓𝒕𝒐𝒕𝒂𝒍 = 𝟎 𝜙𝐸 = (⃗⃗⃗ 𝐸 )(𝐴)(𝑐𝑜𝑠𝜃) Solution = (3.4𝑥1023 )(2.3𝑥10−2 )(𝑐𝑜𝑠84°) Final Answer 𝝓𝒕𝒐𝒕𝒂𝒍 = 𝟎 𝜙𝐸 = 8.17𝑥1020 𝑉 − 𝑚 Final Answer 𝝓𝑬 = 𝟖. 𝟏𝟕𝒙𝟏𝟎𝟐𝟎 𝑽 − 𝒎 GAUSS’S LAW  States that the electric flux through any closed EXAMPLE 2 surface is equal to the net charge inside the What is the electric flux for the following sets of surface divided by the permittivity of free variables? space.  For irregularly shaped objects ⃗⃗⃗ 𝐸 = 9.5𝑥1013 𝑁/𝐶 Given 𝐴 = 1.6𝑥10−5 𝑚2 𝜃 = 75° Asked 𝜙𝐸 = ? Formula 𝜙𝐸 = (⃗⃗⃗ 𝐸 )(𝐴)(𝑐𝑜𝑠𝜃) 𝜙𝐸 = (⃗⃗⃗ 𝐸 )(𝐴)(𝑐𝑜𝑠𝜃) Solution = (9.5𝑥1013 )(1.6𝑥10−5 )(𝑐𝑜𝑠75°) 𝜙𝐸 = 3.93𝑥108 𝑁𝑚2 /𝐶 Final Answer 𝝓𝑬 = 𝟑. 𝟗𝟑𝒙𝟏𝟎𝟖 𝑵𝒎𝟐 /𝑪 fau GENERAL PHYSICS: ELECTROSTATICS ELECTRIC CHARGE | CAPACITANCE | ELECTRICAL FLUX | ELECTRIC POTENTIAL AND POTENTIAL ENERGY SHS | STEM | SECOND SEMESTER | 2024  Electric fields around regularly shaped bodies  Thus, the work done in bringing two charges can be predicted analytically using the closer together from infinity to where they are computations made earlier. separated by the distance r is:  This is because the charges in these regularly 𝒌𝑸𝟏 𝑸𝟐 shaped bodies are uniformly distributed. 𝑾= 𝒓 Qualitative descriptions of the electric field are also possible as the surfaces concerned are  Work is expressed in Joules (J) and it is a symmetrical. scalar quantity. This is also the amount of  The treatment of irregular electric fields electric potential energy stored in the system coming from irregularly shaped surfaces of two charges: requires the use of the differential and integral 𝒌𝑸𝟏 𝑸𝟐 forms of Gauss’s law. An irregularly shaped 𝑼= 𝒓 surface will have varying electric fields at all points of its body. EXAMPLE 1 Two point charges 𝑄 = 32 𝜇𝐶 and 𝑞 = −7𝜇𝐶 are separated by a distance of 19 𝑐𝑚. Find the electric potential energy of the system. 𝑄 = 32 𝜇𝐶 𝑞 = −7 𝜇𝐶 Given 𝑘 = 9𝑥109 𝑁𝑚2 /𝐶 2 𝑟𝑄𝑞 = 19 𝑐𝑚 = 0.19 𝑚 Asked 𝑈=? 𝑘𝑄𝑞 Formula 𝑈= 𝑟 𝑘𝑄𝑞 DEFINITION OF TERMS 𝑈= Solution 𝑟 Energy  Is defined as the 9𝑥10 (32𝑥10−6 )(−7𝑥10−6 ) 9 capacity to do work. = 0.19 𝑚 Potential  Is a form of energy of a Final Answer 𝑼 = −𝟏𝟎. 𝟔𝟏 𝑱 energy body due to its position and normally converted ELECTRIC POTENTIAL into useful work.  Is the amount of electric potential energy per Gravitational  Is a type of potential unit charge. This is equivalent to the amount Potential energy due to a body’s of work needed to move a charge from one Energy elevation from the reference point to another. (GPE) ground.  Electric potential is the amount of electric potential energy per unit charge. ELECTRIC POTENTIAL ENERGY  This is equivalent to the amount of work  Suppose there are two positive charges Q1 needed to move a charge from one reference and Q2 and they are separated by a short point to another. distance r.  Electric potential is mathematically described using this equation: 𝒌𝑸 𝑽𝑬 = =𝑬 𝒓  𝑉𝐸 – electric potential  𝑘 – Coulomb’s constant  Work is required to overcome the repulsion  𝑄 – source charge between the two positive charges to hold  𝑟 – distance from the source charge them together in this way. That amount of  𝐸 – electric field work gives the system of two charges an electric potential energy, U. fau GENERAL PHYSICS: ELECTROSTATICS ELECTRIC CHARGE | CAPACITANCE | ELECTRICAL FLUX | ELECTRIC POTENTIAL AND POTENTIAL ENERGY SHS | STEM | SECOND SEMESTER | 2024 EXAMPLE 1 Direction of Electric Field in Relation to Compute the electric potential from a source Electric Potential charge of +5.02𝑥10−13 𝐶 if a test charge will be  Positive – lower to higher electric potential placed 2.08𝑥10−3 𝑚 from it.  Negative – higher to lower electric potential 𝑄 = +5.02𝑥10−13 𝐶 EQUIPOTENTIAL LINES Given 𝑟 = 2.08𝑥10−3 𝑚 CONSTANT FIELD 𝑘 = 9𝑥109 𝑁𝑚2 /𝐶 2  The electric field lines are perpendicular to the Asked 𝑉𝐸 = ? plates and equipotential lines are parallel to 𝑘𝑄 the plates. Formula 𝑉𝐸 = =𝐸 𝑟 𝑘𝑄 𝑉𝐸 = =𝐸 𝑟 Solution 9𝑥109 (5.02𝑥10−13 𝐶) 𝑉𝐸 = 2.08𝑥10−3 𝑚 𝑽𝑬 = 𝟐. 𝟏𝟕 𝑽 Final Answer 𝑽𝑬 = 𝟐. 𝟏𝟕 𝑽 EXAMPLE 2 POINT CHARGE How much is the electric potential from the  The equipotential lines are circles and a +4.02𝑥10−15 𝐶 source charge if a test charge will sphere centered on the charge is called an be placed 4.55𝑥10−12 𝑚 from it. equipotential surface. The equipotential lines get further apart with increasing radius. 𝑄 = +4.02𝑥10−15 𝐶 Given 𝑟 = 4.55𝑥10−12 𝑚 𝑘 = 9𝑥109 𝑁𝑚2 /𝐶 2 Asked 𝑉𝐸 = ? 𝑘𝑄 Formula 𝑉𝐸 = =𝐸 𝑟 𝑘𝑄 𝑉𝐸 = =𝐸 𝑟 Solution 9𝑥10 (4.02𝑥10−15 𝐶) 9 𝑉𝐸 = 4.55𝑥10−12 𝑚 DIPOLE 𝑽𝑬 = 𝟕. 𝟗𝟒𝒙𝟏𝟎𝟔 𝑽  Mirror symmetry about the center point of the Final Answer 𝑽𝑬 = 𝟕. 𝟗𝟒𝒙𝟏𝟎𝟔 𝑽 dipole  Equipotential lines are everywhere EQUIPOTENTIAL LINES perpendicular to the electric field lines.  Loops drawn around a charge to represent  The plane perpendicular to the line between the electric potential around it. the charges at the midpoint is an equipotential  At any point in the loop, the electric potential plane with potential zero. is constant.  Drawn perpendicular to each electric field line –this means that for every electric field line, there are numerous equipotential lines.  The intersection between the two lines implies the inverse relationship between electric field and the electric potential  In three dimensions, the lines form equipotential surfaces. fau