Theme: Electricity Terms/Laws Definition, Formula and Derivation, Theory/Notes Electric Charge,π Coulomb’s Law π = ππ = ππ = πΌπ‘ = It states that the electrostatic force, π between two separated point charges πΈ and π is directly proportional to the product of the charges, πΈπ and inversely proportional to the square of the distance between them,ππ ππ π2 ππ 1 πΉ=π 2 [π = ] π 4ππ0 Charges in free space, ππ πΉ= 4ππ0 π 2 Charges in insulating medium, ππ πΉ= [π = π0 ππ ] 4ππ0 ππ π 2 ππ πΉ= 4πππ 2 πΉ∝ From Gauss’ law, π πΈ= 4ππ0 π 2 πΉ = πΈπ π πΉ=( )π 4ππ0 π 2 ππ πΉ= 4πππ 2 Electric Field,πΈ A region around a charged particle where free positive test charge will experience an electric force exerted by the particle. π πΈ = π π Repulsive Force - Like charges repel each other with forces that same in magnitude and inopposite direction. - Action-reaction pair that obeys Newton’s Third Law of Motion. Attractive Force -Unlike charged attract each other with forces that same in magnitude and in opposite direction. Electric field produced by : Positive point charge +Q Negative point charge -Q Electric dipole Two like charges Point charge and plate with unlike charges The effect of an electric field on a metal coated polystyrene ball Parallel plates with unlike charges The effects of an electric field on a candle flame Electric Field Strength/Intensity,πΈ It defined as the electric force acting on a unit positive charge placed at the point. Electric force πΉ per unit positive charge +π at the point. πΈ= πΉ π ππ 1 ππ ( ) [πΉ = ] 4ππ0 π 2 π 4ππ0 π 2 π πΈ= 4ππ0 π 2 πΈ= Coulomb’s Law related Motion of a charge in Uniform Electric Field For uniform electric field, πΈ= π π - A force πΉ = πΈπ is acting on a charge+π that causes it accelerates to move in a straight line in the direction od electric field. πΉ = ππ = πΈπ ππ = πΈπ πΈπ πΈπ (πππππππππ‘πππ ππ + π) / π = − (πππππππππ‘πππ ππ − π) π= π π Parabolic path of charge in Uniform Electric Field Gauss’ Law and Electric Flux, π Electric flux π is the number of electric field lines πΈ that intersect/penetrate an area π΄. When A is the normal to E, π = πΈπ΄ When A is not the normal to E, π = (πΈ cos π)π΄ The charge experiences projectile motion. A constant force pulling the charge downward. π£π₯ increases , but π£π¦ remains constant. Gauss’ law states that the electric flux π through a closed surface (Gaussian surface) is given by Σπ π= π0 Electric field produced by : π= Σπ π0 Isolated point charge where Σπ is the sum of the charges enclosed by the surface and π0 is the permittivity of free space. βͺ Electric field strength,E Electric flux through Gaussian surface, π = πΈπ΄ π = πΈ(4ππ 2 ) [π΄ = 4ππ 2 ] Applying Gauss’ law, Σπ π= π0 Q πΈ(4ππ 2 ) = [ Σπ = π] π0 πΈ= π 4ππ0 π 2 Isolated charged conducting sphere βͺ βͺ βͺ Inside the sphere (π < π ), there is no charge. Σπ = 0 If inside the sphere (π < π ) have charges Σπ = π , the mutual repulsive force of like charges would push themselves away from each other until they reach the sphere surface. Hence, charges only appear on the sphere surface and distributed uniformly. β’ Electric field strength inside the sphere (π < π ) Electric flux through Gaussian surface, π = πΈπ΄ π = πΈ(4ππ 2 ) [π΄ = 4ππ 2 ] Applying Gauss’ law, Σπ π= π0 0 πΈ(4ππ 2 ) = [ Σπ = 0] π0 0 πΈ= 4ππ0 π 2 πΈ=0 β’ Electric field strength outside the sphere surface (π ≥ π ) Electric flux through Gaussian surface, π = πΈπ΄ π = πΈ(4ππ 2 ) [π΄ = 4ππ 2 ] Applying Gauss’ law, Σπ π= π0 Q πΈ(4ππ 2 ) = [ Σπ = π] π0 π πΈ= 4ππ0 π 2 On the sphere surface (π ≥ π ), π πΈ= 4ππ0 π 2 Graph E - r Uniformly charged plate Electric Potential,π and Electric Potential Energy,π Electric potential π is the work done π to bring a unit of positive charge +q from infinity ∞ to a point. Electric potential energy π is the work done π to bring the positive charge +q from infinity ∞ to a point. Electric potential and Electric potential energy due to a point charge +Q Force exerted on +q by +Q, ππ 4ππ0 π₯ 2 Work done to bring +q through a small displacement -dx, π = πΉπ ππ = πΉ(−ππ₯) ππ ππ = − ππ₯ 4ππ0 π₯ 2 Work done to bring +q from infinity to a point ,P , πΉ= π π=∫ π π ∞ ππ ππ₯ 2 ∞ 4ππ0 π₯ ππ π π = − [− ] 4ππ0 π₯ ∞ ππ π = −(− − 0) 4ππ0 π ππ π= 4ππ0 π π = −∫ From the definition of electric potential, π = π ππ 1 =( ) π 4ππ0 π π π π= 4ππ0 π Relationship between V and U, π π π = ππ ππ π= 4ππ0 π π= Electric potential due to a charged conducting sphere, β’ Inside the sphere and on the sphere surface (π ≤ π ) π π= 4ππ0 π β’ Outside the sphere (π > π ) π π= 4ππ0 π Graph V-r Equipotential surface A space where all points charge have the same electric potential. π • π = 4ππ 0π at all point. • No work done W in moving a charge from a point to another. π=0 Electric Current,I Rate π‘ of flow of charge Q in a conductor. Current π πΌ= π‘ • • • Current is a form of positive charge flow. Direction of current or direction of positive charge flow is from positive terminal (+) to negative terminal (−). Current can flow is due to a potential difference between two points in a circuit. Ammeter • • Potential Difference / Voltage,π Work done W to bring a unit of positive charge from one point to another against the electric field. π= π= π π Measures current in Ampere (A) It is in series arrangement in a circuit. Potential difference between two points πΈπππππ¦ π‘ππππ πππππ π • • Electric field is a conservative field. Potential difference between two points is same via any path. π1 = π2 = π3 • Velocity of a particle Work done to move +q (electric potential energy), π π= π π = ππ After being accelerated through a potential difference V, Gain in KE = Loss in Electric PE 1 ππ£ 2 = ππ 2 2ππ π£2 = π 2ππ π£=√ π Potential difference between anode and cathode • Velocity of the electron Horizontal component (x-axis) After the electron being accelerated via a potential difference V , Gain in KE = Loss in Electric PE 1 π π£ 2 = ππ 2 π π₯ 2ππ π£π₯ 2 = ππ 2ππ π£π₯ = √ ππ Vertical component (y-axis) Uniform acceleration in the electric field, πΈπ π= ππ When electron emerges from electric field, π£π¦ = π’ + ππ‘ πΈπ πΏ π£π¦ = 0 + ( ) ( ) ππ π£π₯ πΏπΈπ π£π¦ = ππ π£π₯ When electron enter the uniform electric field, π£π₯ constant. • • Time taken to transverse the entire electric field, πΏ π‘= π£π₯ Resultant velocity, v π£ = √π£π₯ 2 + π£π¦ 2 Angle to the x-axis ,π tan π = π£π¦ π£π₯ Voltmeter • • Measures voltage in Volt (V) It is in parallel arrangement in a circuit. Projectile motion- in the electric filed Straight line motion – outside the electric field Relationship between V and E β’ β’ V-x and E-x graphs β’ Uniform electric field in parallel plates Resistance Ohm’s Law Electrical Energy Electrical Power Capacitance,πΆ The ratio of the charge Q on the capacitor to the potential difference V across it. Capacitor • • Used to store and release electric charges and electrical energy. It can be arranged in series and parallel. • It consists of two parallel plates. Charged capacitor • • • • • πΆ= π π When a capacitor is connected directly to a battery, electrons from one plate will be transferred to another. It causes one plate is deficit of electrons with +Q whereas another is excess of electrons with -Q. Two plates have equal amount of charges but different of types. +π = −π The plate with +Q at a higher potential whereas plate with -Q at a lower potential. So, there is a potential difference V across the plates. Graph Q-V # Capacitance is independent with potential difference across it. #π ∝π Capacitance in isolated charged sphere π 4ππ0 π π πΆ= π 4ππ0 π πΆ = π( ) π πΆ = 4ππ0 π π= Capacitance of Parallel-Plate capacitor For uniform electric field, π= π π πππ πΈ = π΄ π Using Gauss’ law, πΈ= π π0 • • Therefore, Then, πΈ=πΈ π π = π0 π π0 π π= π π=π π π0 π = π΄ π π π0 π΄ = π π π0 π΄ πΆ= π • [ π = πΆ] π The electrons on X induce +Q on Y. The more electrons flow to X , the more +Q is induced by electrons , the potential difference V increases. When potential difference across plates = voltage of battery β The flow of electrons from battery stops β The charge on the plates is π = πΆπ Dielectric Series Circuit and Parallel Circuit Series circuit Parallel circuit Diagram πΌ = πΌ1 = πΌ2 = πΌ3 = β― π = π1 + π2 + π3 + β― π = π 1 + π 2 + π 3 + β― π = π1 = π2 = π3 = β― Current Voltage/Potential difference Effective Resistance Charge πΌ = πΌ1 + πΌ2 + πΌ3 + β― π = π1 = π2 = π3 = β― 1 1 1 1 = + + +β― π π 1 π 2 π 3 π = π1 + π2 + π3 + β― π = π1 + π2 + π3 + β― π π π π = + + +β― πΆ1 πΆ2 πΆ3 1 1 1 π = π( + + + β―) πΆ1 πΆ2 πΆ3 π 1 1 1 = + + +β― π πΆ1 πΆ2 πΆ3 1 1 1 1 = + + +β― πΆ πΆ1 πΆ2 πΆ3 • When a bulb/ device is not function in the circuit, the other devices cannot function. Effective Capacitance π = π1 + π2 + π3 + β― π = πΆ1 π + πΆ2 π + πΆ3 π + β― π = π(πΆ1 + πΆ2 + πΆ3 + β― ) π = πΆ1 + πΆ2 + πΆ3 + β― π πΆ = πΆ1 + πΆ2 + πΆ3 + β― Other Features Applications • When a bulb/ device is not function in the circuit, the other devices still can function.
