JEE Class main Answered
⭕ Answer this question with proper Explnation..!
Asked by vidyavikram10 | 13 Mar, 2020, 04:25: PM
Expert Answer
Electric field inside a solid sphere is linear function of radial distance and its value E at surface of sphere is given as
E = Q/ [ 4πεo R2 ]
Where Q is total charge content of sphere, R is radius and εo is permittivity of free space.
In the given graph, Electric field variation with respect to radial distance is linear upto 4 m,
hence radius of solid sphere is 4 m .
From the graph , we get electric field value at R = 4 m as 4 × tan37 = 3.
Hence we consider electric field at surface of sphere as 3 N/C
Hence E = Q/ [ 4πεo × 16 ] = 3 N/C , hence Q = 192 πεo C
Answered by Thiyagarajan K | 13 Mar, 2020, 08:37: PM
Application Videos
JEE main - Physics
Asked by arivaryakashyap | 23 Apr, 2024, 10:40: AM
ANSWERED BY EXPERT
JEE main - Physics
Asked by ratnadeep.dmr003 | 21 Apr, 2024, 11:06: PM
ANSWERED BY EXPERT
JEE main - Physics
Asked by ksahu8511 | 19 Apr, 2024, 11:55: AM
ANSWERED BY EXPERT
JEE main - Physics
Asked by mohammedimroz | 13 Apr, 2024, 09:48: PM
ANSWERED BY EXPERT
JEE main - Physics
Asked by medhamahesh007 | 02 Apr, 2024, 11:11: AM
ANSWERED BY EXPERT
JEE main - Physics
Asked by gundlasumathi93 | 31 Mar, 2024, 02:13: PM
ANSWERED BY EXPERT
JEE main - Physics
Asked by chhayasharma9494 | 31 Mar, 2024, 12:47: PM
ANSWERED BY EXPERT
JEE main - Physics
Asked by archithateja3 | 30 Mar, 2024, 10:23: PM
ANSWERED BY EXPERT
JEE main - Physics
Asked by Machinenineha | 27 Mar, 2024, 05:28: PM
ANSWERED BY EXPERT