2 Sharing of charge & Earthing | Conductors | Electrostatics Class 12 | JEE Mains & Advanced
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4 سال پیش
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This is the second lecture of the Conductor that is a part of the Electrostatics Chapter of class 12 for JEE Mains & Advanced. There are total 4 lectures in one shot for conductor chapter class 12, in which ABJ sir is covering all topics, concepts, examples, and problems that are important for JEE Mains and Advanced.
Topics covered in this Electrostatics (Class 12) Lecture by ABJ sir IITD with timestamp:-
00:00 - 04:05 Properties of conductor: In Electrostatic conditions, the Conductor has equal electric potential at any point. In other words, a conductor is an equipotential surface.
04:05 - 16:03 Electrostatics Problem 1: Based on the concept of the electric potential of a conductor. In this problem, we have a neutral conducting sphere of radius R and a positive charge Q at a distance r from the sphere's center. We have to find the electric potential of the sphere, and also, we have to find potential due to induced charge at different points. ABJ sir solved this problem using the concept of the equipotential surface of the conductor.
16:46 - 25:15 Electrostatics Problem 2: Based on the concept of the electric potential of a conductor. In this problem, we have a conducting neutral sphere of radius R, and a dipole is placed at some angle & at some distance from the sphere's center. We have to find the electric potential of this conducting sphere.
25:16 - 34:15 Sharing of Charge: Sharing of charge is very important for JEE Mains, JEE Advanced & Olympiad. ABJ sir explains this topic with the help of examples. Concept - If two conductors are in contact, the charge will flow from Higher to lower potential until their final potential is equal.
34:59 - 47:32 Electrostatics Problem 3: Based on the concept of charge sharing. In this problem, we have two spheres of radius R1 & R2 with the charge Q1 & Q2. We have to find the final charge on each sphere if they are in contact and also find the energy lost from the system. ABJ sir explained the concept of charge sharing, and we will use the same concept in this problem. The charge will flow from the higher potential to the lower potential until the potential of both spheres becomes equal. We can calculate the difference in the system's potential energy in both cases to find the energy lost.
48:35 - 56:32 Electrostatics Problem 4: Based on the concept of charge sharing. In this problem, we have two conducting concentric shells of radius R1 & R2 with the charge Q1 & Q2. We have to find the final charge on each sphere if they are in contact and also find the heat lost in this process. ABJ sir explained the concept of charge sharing, and we will use the same concept in this problem.
57:08 - 01:08:16 Electrostatics Problem 5: Based on the concept of charge sharing. In this problem, we have three conducting concentric shells of radius R, 2R & 3R with the charge -Q, +Q & +2Q, respectively. We have to find the final charge on each sphere if conducting shell of radius R and 3R is connected. Also, find the value of heat lost in this process. ABJ sir explained the concept of charge sharing, and we will use the same concept in this problem.
01:08:28 - 01:13:33 Concept of Earthing: Electric Potential of earth is assumed to be zero and constant. Whenever we connect a conducting body to earth through a conducting, its potential will get same as earth equals to zero.
01:13:33 - 01:18:15 Example on earthing: In this problem, we have a conducting sphere of radius R and charge Q. At t=0, this sphere is connected to earth. So due to earthing, charge on this sphere will become zero. Also with the help of this example, ABJ sir explains why earthing is done for the home appliances.
01:18:18 - 01:19:50 Electrostatics Problem 6: Based on the concept of Earthing. In this problem, we have two concentric sphere of radius R and r (R is greater than r). If outer sphere is connected to earth. We have to find the value of charge on the outer shell.
01:20:46 - 01:29:49 Electrostatics Problem 7: Based on the concept of Earthing. In this problem, we have three conducting concentric shells of radius R, 2R & 3R with the charge -Q for inner most shell & +Q for the outer most shell, and middle shell is connected to earth wire a conducting wire. We have to find the final charge on each shell after earthing.
Chapter - Conductor & Its properties is very important wrt JEE advanced.
𝐍𝐞𝐰 𝐂𝐨𝐮𝐫𝐬𝐞𝐬 @𝐂𝐨𝐦𝐩𝐞𝐭𝐢𝐬𝐡𝐮𝐧: https://linktr.ee/competishun
𝐂𝐨𝐦𝐩𝐞𝐭𝐢𝐬𝐡𝐮𝐧 𝐖𝐞𝐛𝐬𝐢𝐭𝐞 : https://online.competishun.com/
𝐂𝐨𝐦𝐩𝐞𝐭𝐢𝐬𝐡𝐮𝐧 𝐀𝐩𝐩𝐥𝐢𝐜𝐚𝐭𝐢𝐨𝐧: EduCompetishun (App Link: https://bit.ly/3OJ9vCU)
𝐏𝐂𝐌 𝐂𝐡𝐚𝐩𝐭𝐞𝐫 𝐰𝐢𝐬𝐞 𝐘𝐨𝐮𝐓𝐮𝐛𝐞 𝐏𝐥𝐚𝐲𝐥𝐢𝐬𝐭: https://bit.ly/389Pt22
📞 𝐅𝐨𝐫 𝐪𝐮𝐞𝐫𝐢𝐞𝐬, Contact us at: 8888000021, 7410900901, 7410900906, 7410900908,
#JEEMain #JEEAdvanced #Physics
This is the second lecture of the Conductor that is a part of the Electrostatics Chapter of class 12 for JEE Mains & Advanced. There are total 4 lectures in one shot for conductor chapter class 12, in which ABJ sir is covering all topics, concepts, examples, and problems that are important for JEE Mains and Advanced.
Topics covered in this Electrostatics (Class 12) Lecture by ABJ sir IITD with timestamp:-
00:00 - 04:05 Properties of conductor: In Electrostatic conditions, the Conductor has equal electric potential at any point. In other words, a conductor is an equipotential surface.
04:05 - 16:03 Electrostatics Problem 1: Based on the concept of the electric potential of a conductor. In this problem, we have a neutral conducting sphere of radius R and a positive charge Q at a distance r from the sphere's center. We have to find the electric potential of the sphere, and also, we have to find potential due to induced charge at different points. ABJ sir solved this problem using the concept of the equipotential surface of the conductor.
16:46 - 25:15 Electrostatics Problem 2: Based on the concept of the electric potential of a conductor. In this problem, we have a conducting neutral sphere of radius R, and a dipole is placed at some angle & at some distance from the sphere's center. We have to find the electric potential of this conducting sphere.
25:16 - 34:15 Sharing of Charge: Sharing of charge is very important for JEE Mains, JEE Advanced & Olympiad. ABJ sir explains this topic with the help of examples. Concept - If two conductors are in contact, the charge will flow from Higher to lower potential until their final potential is equal.
34:59 - 47:32 Electrostatics Problem 3: Based on the concept of charge sharing. In this problem, we have two spheres of radius R1 & R2 with the charge Q1 & Q2. We have to find the final charge on each sphere if they are in contact and also find the energy lost from the system. ABJ sir explained the concept of charge sharing, and we will use the same concept in this problem. The charge will flow from the higher potential to the lower potential until the potential of both spheres becomes equal. We can calculate the difference in the system's potential energy in both cases to find the energy lost.
48:35 - 56:32 Electrostatics Problem 4: Based on the concept of charge sharing. In this problem, we have two conducting concentric shells of radius R1 & R2 with the charge Q1 & Q2. We have to find the final charge on each sphere if they are in contact and also find the heat lost in this process. ABJ sir explained the concept of charge sharing, and we will use the same concept in this problem.
57:08 - 01:08:16 Electrostatics Problem 5: Based on the concept of charge sharing. In this problem, we have three conducting concentric shells of radius R, 2R & 3R with the charge -Q, +Q & +2Q, respectively. We have to find the final charge on each sphere if conducting shell of radius R and 3R is connected. Also, find the value of heat lost in this process. ABJ sir explained the concept of charge sharing, and we will use the same concept in this problem.
01:08:28 - 01:13:33 Concept of Earthing: Electric Potential of earth is assumed to be zero and constant. Whenever we connect a conducting body to earth through a conducting, its potential will get same as earth equals to zero.
01:13:33 - 01:18:15 Example on earthing: In this problem, we have a conducting sphere of radius R and charge Q. At t=0, this sphere is connected to earth. So due to earthing, charge on this sphere will become zero. Also with the help of this example, ABJ sir explains why earthing is done for the home appliances.
01:18:18 - 01:19:50 Electrostatics Problem 6: Based on the concept of Earthing. In this problem, we have two concentric sphere of radius R and r (R is greater than r). If outer sphere is connected to earth. We have to find the value of charge on the outer shell.
01:20:46 - 01:29:49 Electrostatics Problem 7: Based on the concept of Earthing. In this problem, we have three conducting concentric shells of radius R, 2R & 3R with the charge -Q for inner most shell & +Q for the outer most shell, and middle shell is connected to earth wire a conducting wire. We have to find the final charge on each shell after earthing.
Chapter - Conductor & Its properties is very important wrt JEE advanced.
𝐍𝐞𝐰 𝐂𝐨𝐮𝐫𝐬𝐞𝐬 @𝐂𝐨𝐦𝐩𝐞𝐭𝐢𝐬𝐡𝐮𝐧: https://linktr.ee/competishun
𝐂𝐨𝐦𝐩𝐞𝐭𝐢𝐬𝐡𝐮𝐧 𝐖𝐞𝐛𝐬𝐢𝐭𝐞 : https://online.competishun.com/
𝐂𝐨𝐦𝐩𝐞𝐭𝐢𝐬𝐡𝐮𝐧 𝐀𝐩𝐩𝐥𝐢𝐜𝐚𝐭𝐢𝐨𝐧: EduCompetishun (App Link: https://bit.ly/3OJ9vCU)
𝐏𝐂𝐌 𝐂𝐡𝐚𝐩𝐭𝐞𝐫 𝐰𝐢𝐬𝐞 𝐘𝐨𝐮𝐓𝐮𝐛𝐞 𝐏𝐥𝐚𝐲𝐥𝐢𝐬𝐭: https://bit.ly/389Pt22
📞 𝐅𝐨𝐫 𝐪𝐮𝐞𝐫𝐢𝐞𝐬, Contact us at: 8888000021, 7410900901, 7410900906, 7410900908,
#JEEMain #JEEAdvanced #Physics
4 سال پیش
در تاریخ 1399/06/01 منتشر شده
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