 वूग जो फ्लुक्स धेए जो वूग ख्लूँट आप देखान, अगर न्चाज जो वह क्यो थ्दबॉन पना वूग भान. तो ए�好啦 as a तो मनुpecially kesh और मैं क् cursory क्sure क्sure क्sure क्sure क्sure क्sure क्sure थावुझा मानेallas क्sure क्sure क्sure क्sure इस आप मुझा मैं कि थाद रवुझा एँ मैं क्sure उतके ले बार आ� plank's is proportional to how many charges in closed right then i can write this as flux some constant let us say c is constant c into q enclosed right now i want value of constant right to know the constant value consider a case if you have point charge ijuana wią Extremadura ॐ ॐ ॐ ॐ ॐ ॐ ॐ ॐ ॐ ॐ ॑ ॐ constant theta to cos theta will be 1 or electric field is constant along this entire line and that is equal to what if distance is r the electric field is kq by r square even electric field magnitude is constant so when you try to find out flux then what will happen e into a into cos of 0 will happen right so e into a will happen and how much kq is r square area is 4 r square okay now if I expand k to 1 by 4 pi islam knot in this way if I write 4 pi and 4 pi is cut r square r square cut this will come 1 by islam knot times q okay so constant value you have taken 1 by islam knot although you have taken a special case but constant value should not change depending on what type of case you are taking fine so that 1 by islam knot will be so basically now we have taken the fact that the flux is q enclosed by islam knot and the flux is the integral over closed surface for a dot d a so this will be q enclosed by islam knot right so this type of formulae can be done fine so this is a very important law all right and using this law we will now take out the electric field of different cases okay now we will use the Gauss law to take out the electric field okay now to use the Gauss law we just discussed flux is integral of 3 variables so we will take out the integral from 3 variables and we will equate it to q enclosed by islam knot and we will take out the electric field from there now when can you take out the electric field when you take out the integral and when can you take out the integral when you have discussed those two special cases they will satisfy okay and on the left hand side you should have only one variable so the electric field should be constant whatever you take out the surface right so we will take out the Gauss law first of all first of all we will know that how to take out the Gauss surface okay so write Gauss surface so it is an imaginary enclosure it need not be real okay so it is an imaginary enclosure it should not cut discrete charges so if there is a point charge lying somewhere that should not cut across by the Gauss surface all right it may cut the it can cut the continuous charge distribution fine now these are three things these are must fine these three things without the Gauss surface can not be imagined these three have to happen and if the other things that I am going to say will satisfy then you can easily take out the electric field okay so these three are must to have and good to have is this electric field magnitude is constant and theta is constant over surface fine or theta is 90 degree fine so if this thing will happen these two things something will happen then you easily take out the integral and it may be that you take out the electric field very easily okay so this is the theory after taking out the electric field practically let's see how the electric field comes okay first of all we will apply Gauss law in symmetrical cases okay and there are three types of symmetry which are spherical symmetry, cylindrical symmetry and planar symmetry okay so symmetry number one is spherical symmetry right understand that you have a solid sphere here a solid sphere this may volume charge density rho fine and you have to take out the electric field at a distance r its radius is capital r okay this case we have already integrated and we have already taken it out but now we will try to take it out by applying Gauss law we want something electric field at a distance r so Gauss surface should pass through distance at a distance r otherwise when we apply Gauss law then the electric field comes out on the surface so if the surface does not pass through this point then the electric field will not come in the equation okay so it is very important to pass through this point of the surface so I will make this surface symmetrically fine and here I will write it as Gauss law now integral e.da is equal to q enclosed by epsilon not fine now this dotted line you are seeing this is my Gaussian surface in this Gaussian surface I know that electric field will be like this and area vector will be like this so the angle between the two is zero right and the magnitude of the electric field is constant okay so this left hand side what will happen is e.da into cos of zero because zero angle is throughout and e is constant throughout so e and cos of zero is outside cos of zero is one into integral of da this will be simply e into da integral of four pi r squared okay and how much is q enclosed q enclosed is your volume charge density into this thing's volume because in this volume the charge is enclosed right so q enclosed will be rho into four by three pi r cubed okay so if I substitute one or two from here then I will get equation like this e into four pi r squared is equal to rho into four pi r cubed by three epsilon not so e into four pi r cubed by three epsilon not so this way I have removed the magnitude of electric field I should know the direction first the direction is radially outward fine for a positive rho for negative rho when we put minus automatically the direction of electric field will be reversed so in this way you can remove electric field in the case of spherical symmetry okay you will remember one question we asked in which we took atomic model we took z times e charge at the center and we considered electron as a spread in a density on radius r fine and we take radius capital R and we had to remove electric field at a distance small r which is less than capital R fine so let's try this with Gauss law to solve fine so along with this when I try to remove the integral which is my flux so this will be e into four pi r squared this is simple right and if I want to remove q enclosed so how will I remove it one positive charge is enclosed that is a point charge its value is z times e and then the electron density is also enclosed how much its density is sorry how much its charge is minus z e r by r whole q how does it come the whole atom is neutral so z e is the whole volume four pi r cubed so the volume charge density will be e divided by 4 by 3 pi capital R cubed into 4 by 3 pi small r cubed will be how much its charge is inside so that will come fine so now the flux which is enclosed by islam knot is equal to so e into 4 pi r squared is equal to z e 1 minus r cubed by capital r cubed divided by islam knot right so from here my electric field comes z e by 4 pi epsilon knot 1 by r squared minus r by r cubed fine so in this way you can remove the electric field of that particular case by applying Gauss law ok now we will talk about a different type of symmetry which is cylindrical symmetry see cylindrical symmetry means that if you keep a cylinder then everything will be uniform on the lateral surface area ok so this symmetry is found in an infinite wire this is an infinite wire with line charge density lambda fine and you have to remove electric field at a distance r so imagine a Gauss surface here this is the cylinder this is the cylinder ok you can consider this cylinder height h so my enclosure here is a cylinder fine and there are two surfaces of the cylinder surface 1 lateral surface which is surface 2 and another surface surface 3 ok so the flux integral e dot dA I divide this into three parts surface 1 e dot dA plus surface 2 e dot dA plus surface 3 e dot dA fine now the direction of the area vector is the direction of the area vector for surface 1 for surface 2 and surface sorry for surface 3 and surface 2 is the lateral surface so it will be circular radially outward fine so if you see from the front then some area vector will be right and where will be the direction of the electric field this is an infinite wire and we have seen the infinite wire that e y is 0 and only e x is there e x means along the perpendicular of the wire fine so the electric field will also be like this in this direction right so here also the electric field will be like this now tell me what is the angle of the electric field and area vector here 90 here also 90 and here also 0 fine and if 90 degree is done my friend so can you say that this is 0 and this is also 0 you can say it so net net I have this is the whole integral that is just equal to integral of surface 2 e into dA into cos of 0 which is 1 ok now you will see that in lateral surface area all the points are equidistant from the wire right understand this distance is small r fine so because of symmetry the magnitude of the electric field along the lateral surface becomes constant so the integral this will be e this and we will write this further integral of dA is what total area which is lateral surface area so this will be e into 2pi r into height h alright and this flux is equal to q enclosed by islam knot right so q enclosed by islam knot how much is this lambda into h is q enclosed lambda is what charge period length alright divided by islam knot this is equal to e into 2pi r h right so this will give me electric field as lambda divided by 2pi islam knot into r so this is the same electric field we integrated very difficultly we took out lambda by 2pi islam knot r it is very easy to get goslo but goslo you can get just an infinite wire because that only gives you a symmetrical case alright so this is cylindrical symmetry let's talk about one last symmetry that is planar symmetry alright write planar symmetry comes from infinite sheet so this is your infinite sheet so when you go to infinite sheet so electric field will come out so electric field will come out like this okay so this is from here also like this you will get 2d surface that infinite wire but it's not like all this electric field is going like this and from here is going like this it's not circular like in wire alright so you understand this and these two lines are parallel all the lines parallel and are going like this so distance will not get then electric field is uniform electric field is uniform so it should be independent of distance electric field per unit area sorry lines per unit area which is electric field is not changing so that is uniform okay so come in this case assume this is your Gaussian surface this is surface area this is A understand again this you understand this is H okay so this is some area from surface is cutting okay now I have to integral E.da to get this first surface second surface third surface this will be equal to integral over first surface E.da plus over second surface E.da third surface E.da okay so you will know that this area vector is from surface it is out of and electric field is like this how is the angle from between electric field and area vector it is a 90-degree so this second surface is across this integration is 0 okay and this तो यी जो आएगा मेरे दोस, वो आएगा सिग्मा बाई 2 इप्सलम नोद. तो इस तरे से हम ने एलेक्टी फिल निकाल ली एक इन फाइनाइट शीट के ड़ारा, जिसका सर्फिस जाज जेंस ती सिग्मा है. तो इस तरे से तीनो सिमिट्री तेकन केर अफ हो जाती हैं, अगर आप गोस्लो लगा हैं तरीके से. अप चलिए अब हम ने गोस्लो तो पडी लिया है, तो उसके उपर एक आद सवाल लगा लिते हैं. अप आप ये छीज दियान देखही है, हम लोग ये सब हम कुन्षे अब वीटिटियो आ अपने है। और अप यो साड, वीटियो के बाद हम्रे जेंई मेईज लेवल के प्रूलम्स होगे, उंबना तो आप बचले लिएक देखान में जैय मेंज लेएल के प्रड़ूम्स के बाभें दीकशन करेंगे यागा और अमारा जैयदवांस लेएल पे भी बचले प्रूम लोँ याँ चप्रेड़ विडियो में डीसकरेंगे तो ये अभी एक सवाल ले ले ले थें जो अभी तो ठौर ना सब खॉत अप, ने वो, गरसा, या उप अउप उग़नो, अगऱ्े, या औद्वा दौप, धौर करेशर, आगरो डिسचते दौब उगय की थी जोई, अगऱ्े वो बग़गे लीगरे आस्रना कोगा, तो जो ख़वा एक प्रिएव लेगरे आर्वो देखॉ� ह Kennedy k k एकि एक भ seventy ठु़ेेÉéeréééééééééééééééééééééééééééééééééééééé éaan आग of य Essentials हुह उठी इल आप ती शाम न कॉन्से printing कॉ अगद है 3 Thank you तो सर्फ दो सर्फेस हैं जोकी एक सर्फेस वनेटी दिग्री आंगल बनार हैं, तुस्रा सर्फेस जिरो दिग्री आंगल बनार हैं. रेस्त नो दे सर्फेस is having any flux. तीके, because every other surface other than these two are making 90 degree angle. तो सर्फ हम ये तो सर्फेस हैं, सर्फेस वन और सर्फेस तु, जिसको हम कनचीटर करेंगे. तो चल्ये निकालते हैं, flux will be equal to total surface pay E.da. तीके, जोकी एकल हो जाता है, सर्फेस वन पे एड़ दीए, plus surface 2 pay E.da. तो क्यो, क्यो, क्यों बाह्ती सर्फेस हैं उस पे E.da 0 हो रहा है, angle 90 degree है. तो E.da लिव्टन साईट पे कितना होगा? ए तो कोस्संट ये. लेठ वाज़ी हैं, ती कोसाआआ वन जीड़गारी, तो मैझने ए जायागा एन अंटिकरल आप दीए. यौगयोगया यह सर� ships 1 पे लेक्टीख्लिट यह वह सर्झे стूब आए सर्फेस 2 पे लेक्टीख्लिट आए ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ ौ  तो, एद्व के बैली होँगाएगी। जब अप औब इस एकस कुछ कर आएड़ालेंगे। तो, ये होझाएगा। एद्व येस अलफ बाप रूट 2 अग, यास्वान्स वीघ इन्स्फार तो तो तो तो तो तो तो thats right!