JavaScript is disabled on your browser. Enable JavaScript to use this site. Learn more

Animated Physics materials overview (XIIth)

Add to Favourites

Post to:

Join the PMT Medical Entrance Exam Community

Description
Animated Physics materials for IIT-JEE, AIEEE, AIPMT, SAT & Boards

Subscribe complete Physics study materials from http://www.vidyadrishti.com/phypackage.php
Mail: vidyadrishti@gmail.com Phone: 09333377572

Comments

Presentation Transcript

Slide 3 : Coulomb’s Law © P K Bharti, IIT Kharagpur www.vidyadrishti.com C 7e 7p 15e 15p 3e A B 4e 7p 18e 15p A B INITIAL STATE FINAL STATE

Charge Property 3 (Quantization) : Charge Property 3 (Quantization) The charge is quantized. (IMPORTANT) It means the total charge Q carried by a charged body is an integral multiple of charge e on an electron. Therefore, we can write, Q = ne where n = an integer = …,-3, -2, -1, 0 , 1, 2, 3,… MEANING OF QUANTIZATION: Quantization is simply step by step. It is not continuous. In the case of charges, the step size is charge on an electron. The charge on a body can be integral multiple of charge on electron (e). Therefore, charge on a body can be 2e, -100e, 576e. But charge on a body can’t be 2.7e, because 2.7 is not an integer. Clearly, the least amount of charge on a body will be ± 1e . NOTE: Please note that negative charge is not less than positive charge. E.g., you cannot say that -1e < 1e. Negative and positive sign of charges simply tell that they are of opposite nature. For historical reasons, charge on an electron is taken as negative and charge on a proton is taken as positive. It could have been taken in opposite manner also. But we follow the historical rule. e 0 1e 2e -2e -1e 4 P K BHARTI, IIT KHARAGPUR

Slide 5 : Two particles A & B, each carrying charge Q, are held fixed with a separation d between them. A particle C having mass m and charge q is kept at the middle point of line AB. (a). If C is displaced through a distance x perpendicular to AB, what would be the electric force experienced by it. (b). Assuming x<

Slide 6 : ELECTRIC FIELD © P K Bharti, IIT Kharagpur www.vidyadrishti.com E E q d F E q x R

Example 7 : Example 7 Q. A circular wire loop of radius a carries a total charge Q distributed uniformly over its length. A small length dL of the wire is cut off. Find the electric field at the centre of the due to the remaining wire. SOLUTION: Applying the concept of previous solution, net electric field at the centre of the circular wire before cut = zero. (because charge distribution is uniform). Thus, field (E) due to the remaining portion of the wire is equal and opposite to that of the E because of cut part. Thus, now charge per unit length of the wire = Q/ 2p a Therefore, charge on length dL (cut), q = (Q / 2 p a) dL Hence, field (E) due to the remaining portion of the wire at the centre = electric field at the centre of the wire because of cut part P K BHARTI, IIT KHARAGPUR 7 dL E E

Slide 9 : ELECTRIC POTENTIAL Example 1: Point Charge METHOD 1: For a single charge Q, the potential is given by V = This shows that V is a constant if r is constant . Thus, equipotential surfaces of a single point charge are concentric spherical surfaces centered at the charge. METHOD 2: We know that electric field lines are in the radial direction for a point positive charge. Since electric field is perpendicular to equipotential surface at each point, therefore equipotential surfaces will be concentric spheres as shown in the figure. Equipotential Surface Equipotential surfaces Equipotential surfaces Electric Field line

Slide 13 : Let us assume the electric field and potential difference between the plates before the dielectric is inserted are respectively. When a dielectric material is inserted between the plates of a capacitor (keeping the charge constant) the electric field and hence electric potential difference decreases by a factor K. This K is known as dielectric constant of that material. Electric field is decreased because induced charge of opposite sign +Qi and –Qi appears on surfaces of the dielectric. These induced charges produce electric field inside the dielectric in opposite directions and as a result net electric field is decreased by a factor K. Clearly, Effect of Dielectrics CAPACITORS +Q – Q +++++++ +++++ ___________ E +++++++ +++++ ___________ E E0 Ei +Qi –Qi

Slide 14 : Case 2: Dielectric slab partially fills the gap between the plates In this case, E0 is the electric field inside the region where dielectric is not present. The electric field inside the region where dielectric is present is E= E0 /K from the definition of dielectric constant. We can reconstruct this capacitor as shown. Hence, the potential difference between the plates of the capacitor is Hence the capacitance is CAPACITORS Parallel plate capacitor with a dielectric +Q – Q E0 +Q – Q E0 E0 t d t d

Slide 16 : Direction of current The charges passing through the surface can be positive or negative, or both. It is conventional to assign to the current the same direction as the flow of positive charge. In electrical conductors, such as copper or aluminum, the current is due to the motion of negatively charged electrons. Therefore, when we speak of current in an ordinary conductor, the direction of the current is opposite the direction of flow of electrons. SI unit of current The SI unit of current is the ampere (A). Clearly, from definition of current, Hence, 1 A is defined is the current when 1 C of charge passes through a cross section in 1s. Current Electric Current Direction of flow of positive charge Direction of flow of current Direction of flow of negative charge (electrons) Direction of flow of current

Slide 18 : Example 6: At first look it seems that resistance are connected in parallel. Idea is that you can stretch a wire with your hand. If you stretch this combination from both hands, the system will look as shown in figure. Clearly the three resistors are in series. Moral: You can stretch or modify a circuit into a simplified form. Electric Current Combination of Resistors

Slide 20 : Hint: Here all resistors are connected between points A and B. As each resistance is equal to r, the entire circuit is symmetrical about line AB. Therefore all points lying on perpendiculars drawn to AB are at the same potential. Thus points (C, D), (E, F, G) and (H, I) have same potential. Thus, we can connect points (C, D), (E, F, G) and (H, I) together as shown. Remaining circuit is a simple series parallel circuit. You can work out yourself to find out the equivalent resistance between A and B. Electric Current Exercise 2 D E F G H I C

Slide 21 : Method 4: Connection removal method This method is useful when the flow of current is a mirror image between input and output above a particular axis. In such cases some junctions are unnecessarily made. Even if we remove that junction there is no difference in the remaining circuit or current distribution. Example 6: We are going to find out the equivalent resistance between points A and B in the given circuit using connection removal method. Current in all branches are shown in figure using symmetry. Same currents are shown with same colour. (How? Think yourself. Don’t skip.) Clearly the circuits is symmetrical about the dotted line. Since same current i4 flows through both middle resistors, we can see that the junction where i2 and i4 are meeting can be removed easily and we can redraw the circuit as shown. This does not affect the original circuit. Electric Current Use of symmetry in circuits A B

Slide 23 : When an ideal voltmeter is connected across an element, the potential difference read by it is the potential difference across that element. Voltmeter must be connected in parallel to that element. STEPS: Step 1: Open-circuit the voltmeter. It means remove the voltmeter from the circuit. Step 2: Find the potential difference across the required element. Voltmeter reading will be same. For example, let us consider the given circuit. Suppose an ideal voltmeter is connected across 2 W resistor. Let us follow the various steps to find out the reading of this ideal voltmeter. Step 1: Voltmeter is open-circuited, i.e., removed from the circuit. Step 2: The current in the branch BC is 6A. Therefore potential difference across 2 W resistor = iR = 6 x 2 = 12 V. Hence, the potential difference read by the voltmeter = potential difference across 2 W resistor = 6 V. Application Method Electric Current 6 A 11 A V

Slide 24 : Q. The resistances of the ammeter and voltmeter are 2.0 W and 200 W respectively. Determine the readings of the voltmeter V and the ammeter A. SOLUTION: Note that the ammeter and voltmeter are not ideal. Therefore we should replace ammeter and voltmeter with their respective resistances. The resulting circuit is a simple series parallel circuit. Therefore, current in the circuit is given by Example 7 Electric Current 50 W i V 200 W d c V 40 W 2 W 2 W V i i

Slide 25 : The potentiometer is an instrument that can be used to measure the emf of a source without drawing any current from the source; it also has a number of other useful applications. Potentiometer consists of a long resistance wire, usually 5 to 10 m long. The wire has a uniform cross-section. Hence, resistance of this wire is directly proportional to the length of the wire. Also from Ohm’s law V = iR. Hence, the potential difference across the wire is directly proportional to the length of the wire. Potentiometer Electric Current Potentiometer Circuit Symbol for Potentiometer

Slide 27 : Magnetic field due to a current Direction of Magnetic Field : Recalling how we arrived at the direction of Vector product of two vectors , Vector direction of the current element dl( i.e. direction of the current) Sweep with your Right hand from to . The direction of your thumb gives the direction of the cross product, Magnetic field in this case Considering at another point Q Coming out the slide, i.e. computer (coming on to you) Going into the slide, going away from you MAGNETIC FIELD

Slide 28 : Magnetic field due to current in a straight wire A simple rule to find the direction of magnetic field due to a current carrying wire A slight variation of right hand thumb rule Stretch the right hand thumb along the direction of the current and curl the fingers to pass through the point P Direction of the fingers gives direction of the magnetic field at that point. While curling the fingers come out of the plane of the slide towards you in this case. MAGNETIC FIELD

Slide 30 : Now let the charge q be given a velocity V and again observed as before to see it behaves A SIMPLE EXPERIMENT CHARGE GETS DEFLECTED TOWARDS THE CURRENT CARRYING WIRE Result : Current carrying wire Charge q A i Velocity V MAGNETIC FIELD

Slide 31 : Quick review Guess the direction of the force… Note: Since q is – ve direction of force is in the opposite direction of the thumb(direction of vector cross product) Direction of Vector Cross Product MAGNETIC FIELD

Slide 33 : A SMALL EXPERIMENT Let us consider a magnet along the axis of the loop. Axis of the loop Now let us move the magnet towards the loop. Result : pointer in the galvanometer deflects while the magnet is in motion Deflection indicates presence of current in the loop Pointer comes back to zero position once the magnet stops moving It means no current in the loop if magnet stops N S ELECTRO-MAGNETIC INDUCTION

Slide 35 : Bohr’s model Bohr’s Atomic Model Diagrammatical representation of series for hydrogen atom n = 1 n = 2 n = 3 n = 4 n = 8 E1 = -13.6 eV E2 = -3.4 eV E3 = -1.5 eV E4 = -0.85 eV E = 0 eV Lyman Series Balmer Series Paschen Series

Slide 36 : PHOTOELECTRIC EFFECT & WAVE PARTICLE DUALITY © 2007-08 P K Bharti, IIT Kharagpur www.vidyadrishti.com PHOTOELECTRIC EFFECT cathode anode Light rays electron Co-author: P Rohit Gandhi IIT Kharagpur

Slide 37 : Photo electric effect PHOTOELECTRIC EFFECT Let us conduct a small experiment. Let there be a metal surface. Let light of sufficient wavelength fall on the surface. Electrons being emitted from the surface of the metal.

Slide 39 : Hints : current logs voltage by p/2 in purely inductive circuit. 2.(c) Voltage leads current by p/2 in a purely inductive circuit.

Slide 40 : An L-C-R series circuit with 100? resistance is connected to an ac source of 200V and angular frequency 300 rad/s. When only the capacitance is removed, the current lags behind voltage by 600 . When only the inductance is removed, the current leads the voltage by 600. Calculate the current and the power dissipated in the L-C-R circuit. Solution : When capacitance is removed, resulting circuit is an RL series circuit. Therefore, When inductance is removed, resulting circuit is an RC series circuit. Therefore,

Slide 41 : Suppose a spherical wave of light is travelling along the radial direction from origin O. We can assume a point source at origin O. Clearly, light wave will propagate in a radial direction. The amplitude is inversely proportional to radial position r from the origin O for a spherical wave of light. The equation of such a light wave may be written as (spherical wave of light travelling along radial direction) Waves Spherical Light Waves

Slide 42 : Wave fronts of a spherical wave If we have a point source emitting waves uniformly in all directions, then the locus of points which have the same amplitude and vibrate in the same phase are spheres and we have what is known as a spherical wave as shown in Fig. Wave fronts of a plane wave At a large distance from the source, a small portion of the sphere can be considered as a plane and we have what is known as a plane wave as shown in Fig. Waves Wave front v Spherical wave fronts Plane wave fronts

Slide 43 : Paragraph for Qs. 1 to 3 The figure shows a surface XY separating two transparent media, medium-1 and medium-2. The lines ab and cd represent wavefronts of a light wave travelling in medium-1 and incident on XY. The lines ef and gh represent wavefronts of the light wave in medium-2 after refraction. 1. Light travels as a (A) parallel beam in each medium (B) convergent beam in each medium (C) divergent beam in each medium (D) divergent beam in one medium and convergent beam in the other medium. 2. The phases of the light wave at c, d, e and f are fc, fd, fe and ff respectively. It is given that fc ? ff . (A) fc cannot be equal to fd (B) fd can be equal to fe (C) (fd - ff) is equal to (fc - fe) (D) (fd - fc) is not equal to (ff - fe) 3. Speed of the light is (A) the same in medium-1 and medium-2 (B) larger in medium-1 than in medium-2 (C) larger in medium-2 than in medium-1 (D) different at b and d Example (IIT-JEE 2007)

Slide 45 : X-rays are produced whenever high-speed electrons collide with a metal target. Any X-ray tube must therefore contain a source of electrons, a high accelerating voltage, and a metal target. Furthermore, since most of the kinetic energy of the electrons is converted into heat in the target, the latter must be water-cooled to prevent its melting. Source of electrons (Cathode) Metal Target (Anode) Electron beam X-Ray beam Coolidge Tube High Accelerating Voltage Evacuated glass tube

Slide 46 : We are studying about reflection of eight from Std. VIth. Let us revise laws of reflection once again. Laws of reflection : Angle of incidence = angle of reflection i.e., L i= L r. The incident ray, the reflected ray and normal to the surface at the point of incident, all lie in the same plane. Normal Reflecting surface Incident ray r i Reflected ray The incident ray, reflected ray and the normal to the reflecting surface lie in the same plane.

Slide 47 : We are going to study reflection from two kinds of mirror – Plane Mirror and Spherical Mirror. Spherical mirrors are further classified in Concave mirror and Convex Mirror. All mirrors are shown below. Reflecting surfaces of all mirrors are shown in sky blue. Mirror Plane mirror Spherical mirror Convex mirror Concave mirror

Slide 48 : To make an image we need at least two reflecting rays. Let us first study about image formation in a plane mirror. Case 1: Image of a point object placed in front of a plane mirror: Let us consider a point object O be placed in front of a plane mirror MM’. To make image of a point object O, we will draw two rays originating from O. 1st ray: Let us consider a ray OA, which incidents normally on mirror MM’. Clearly it will retrace its path along OA after reflection. 2nd ray : Let us again consider a ray OB which incidents at some angle i to the normal of MM’. Clearly if will follow path BC at angle i after reflection. Image of an object lies at a point where two reflected rays, meet or ‘appear to meet’. For this case, two reflected rays never meet. What to do? Yes! We can produce AO and BC in back side of mirror so that these can appear to meet at point O’. We have shown these as dotted lines. Thus the image of point object O forms at point O’ behind the mirror. Note that the two reflected rays ‘appears’ to meet at point O’, therefore, image O’ of object O is not real. This image is imaginary. If the two reflected rays meets at a point, then the image is real. i i O A B M O’ M’ C

Slide 49 : Like reflection, we need to draw at least two refracting rays for image formation (obviously using laws of refraction!). If the two refracted rays actually meet at a point, the image is real. If the two refracted rays appear to meet at a point, the image is virtual or imaginary. Let us form image of a point object after refraction from a plane surface. We will consider four cases. Case 1: Point object O is placed in a denser medium We take two rays as shown: Rarer medium Denser medium i i r r O D C B A First ray: Incident ray OA normal to the surface. Refracted ray AD passes normally (undeviated). Second ray: Incident ray OB at an angle i with normal to surface. The ray of light is travelling from denser to rare medium and hence, refracted ray BC bends away from the normal at an angle r. Clearly, two refracted rays (blue rays) never meet. But they appear to meet at point O’ when produced. Therefore, O’ is the virtual image of object O. Surface O’