UPSC Exam > UPSC Questions > Electric field intensity at point B due to a ...
Start Learning for Free
Electric field intensity at point B due to a point charge Q kept at point A is 24 N/C and the electric potential at point B due to the same charge is 12J/C.Calculate the distance AB and the magnitude of charge Q.?
Most Upvoted Answer
Electric field intensity at point B due to a point charge Q kept at po...
Calculating Distance AB and Charge Q
Electric field intensity is given by the equation E = kQ/r^2, where k is the electrostatic constant, Q is the charge, and r is the distance between the point charge and the point where the field is being measured.
Finding the Distance AB
Given that the electric field intensity at point B due to the charge Q at point A is 24 N/C, we can write:
24 = kQ/AB^2
Similarly, the electric potential at point B due to the charge Q at point A is given by V = kQ/AB.
Given that the electric potential at point B is 12 J/C, we have:
12 = kQ/AB
Dividing the two equations, we get:
24/12 = AB^2/AB
AB = 2
Therefore, the distance AB between point A and point B is 2 units.
Finding the Magnitude of Charge Q
Now that we know the distance AB, we can find the magnitude of charge Q by substituting back into one of the equations. Let's use the equation for electric potential at point B:
12 = kQ/2
Q = 24/k
Since k is a constant (8.99 x 10^9 Nm^2/C^2), we can calculate the magnitude of the charge Q:
Q = 24/(8.99 x 10^9)
Q ≈ 2.67 x 10^-9 C
Therefore, the magnitude of the charge Q is approximately 2.67 x 10^-9 C.
Electric field intensity is given by the equation E = kQ/r^2, where k is the electrostatic constant, Q is the charge, and r is the distance between the point charge and the point where the field is being measured.
Finding the Distance AB
Given that the electric field intensity at point B due to the charge Q at point A is 24 N/C, we can write:
24 = kQ/AB^2
Similarly, the electric potential at point B due to the charge Q at point A is given by V = kQ/AB.
Given that the electric potential at point B is 12 J/C, we have:
12 = kQ/AB
Dividing the two equations, we get:
24/12 = AB^2/AB
AB = 2
Therefore, the distance AB between point A and point B is 2 units.
Finding the Magnitude of Charge Q
Now that we know the distance AB, we can find the magnitude of charge Q by substituting back into one of the equations. Let's use the equation for electric potential at point B:
12 = kQ/2
Q = 24/k
Since k is a constant (8.99 x 10^9 Nm^2/C^2), we can calculate the magnitude of the charge Q:
Q = 24/(8.99 x 10^9)
Q ≈ 2.67 x 10^-9 C
Therefore, the magnitude of the charge Q is approximately 2.67 x 10^-9 C.
Attention UPSC Students!
To make sure you are not studying endlessly, EduRev has designed UPSC study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in UPSC.
|
Explore Courses for UPSC exam
|
|
Top Courses for UPSCView all
Electric field intensity at point B due to a point charge Q kept at point A is 24 N/C and the electric potential at point B due to the same charge is 12J/C.Calculate the distance AB and the magnitude of charge Q.?
Question Description
Electric field intensity at point B due to a point charge Q kept at point A is 24 N/C and the electric potential at point B due to the same charge is 12J/C.Calculate the distance AB and the magnitude of charge Q.? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about Electric field intensity at point B due to a point charge Q kept at point A is 24 N/C and the electric potential at point B due to the same charge is 12J/C.Calculate the distance AB and the magnitude of charge Q.? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Electric field intensity at point B due to a point charge Q kept at point A is 24 N/C and the electric potential at point B due to the same charge is 12J/C.Calculate the distance AB and the magnitude of charge Q.?.
Electric field intensity at point B due to a point charge Q kept at point A is 24 N/C and the electric potential at point B due to the same charge is 12J/C.Calculate the distance AB and the magnitude of charge Q.? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about Electric field intensity at point B due to a point charge Q kept at point A is 24 N/C and the electric potential at point B due to the same charge is 12J/C.Calculate the distance AB and the magnitude of charge Q.? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Electric field intensity at point B due to a point charge Q kept at point A is 24 N/C and the electric potential at point B due to the same charge is 12J/C.Calculate the distance AB and the magnitude of charge Q.?.
Solutions for Electric field intensity at point B due to a point charge Q kept at point A is 24 N/C and the electric potential at point B due to the same charge is 12J/C.Calculate the distance AB and the magnitude of charge Q.? in English & in Hindi are available as part of our courses for UPSC.
Download more important topics, notes, lectures and mock test series for UPSC Exam by signing up for free.
Here you can find the meaning of Electric field intensity at point B due to a point charge Q kept at point A is 24 N/C and the electric potential at point B due to the same charge is 12J/C.Calculate the distance AB and the magnitude of charge Q.? defined & explained in the simplest way possible. Besides giving the explanation of
Electric field intensity at point B due to a point charge Q kept at point A is 24 N/C and the electric potential at point B due to the same charge is 12J/C.Calculate the distance AB and the magnitude of charge Q.?, a detailed solution for Electric field intensity at point B due to a point charge Q kept at point A is 24 N/C and the electric potential at point B due to the same charge is 12J/C.Calculate the distance AB and the magnitude of charge Q.? has been provided alongside types of Electric field intensity at point B due to a point charge Q kept at point A is 24 N/C and the electric potential at point B due to the same charge is 12J/C.Calculate the distance AB and the magnitude of charge Q.? theory, EduRev gives you an
ample number of questions to practice Electric field intensity at point B due to a point charge Q kept at point A is 24 N/C and the electric potential at point B due to the same charge is 12J/C.Calculate the distance AB and the magnitude of charge Q.? tests, examples and also practice UPSC tests.
|
|
Explore Courses for UPSC exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup