Consider a sphere of radius r having charge q unifotmly distributed inside it.ay wgat minimum disyance from its surface the electric otential be half
Answers
Answered by
61
Final Answer : R/3
Steps and Understanding:
1) Given :
Solid sphere is uniformly charged with charge Q.
Asked : Minimum distance from its surface such that Electric potential is half of that in its centre.?
2) We know that,
Electric field of sphere at distance( r) from centre is given by :
E = KQ/R^3 ( 3R^2 - r^2) when r < = R.
E = KQ/r. when r > R.
where
r = distance from centre.
R = Radius of sphere.
3) Electric field at centre at r =0 ,
E(c) = 3KQ/2R.
Half of it,
E(h) = 3KQ/4R.
4) We observe that, E(h) lies at r > R as shown in graph.
Therefore,
E(h) = KQ/r
=> 3KQ/4R = KQ/r
=> r = 4R/3 .
Therefore, distance from centre is 4R/3.
Distance from surface, d = 4R/3 - R = R/3 .
---------------
Steps and Understanding:
1) Given :
Solid sphere is uniformly charged with charge Q.
Asked : Minimum distance from its surface such that Electric potential is half of that in its centre.?
2) We know that,
Electric field of sphere at distance( r) from centre is given by :
E = KQ/R^3 ( 3R^2 - r^2) when r < = R.
E = KQ/r. when r > R.
where
r = distance from centre.
R = Radius of sphere.
3) Electric field at centre at r =0 ,
E(c) = 3KQ/2R.
Half of it,
E(h) = 3KQ/4R.
4) We observe that, E(h) lies at r > R as shown in graph.
Therefore,
E(h) = KQ/r
=> 3KQ/4R = KQ/r
=> r = 4R/3 .
Therefore, distance from centre is 4R/3.
Distance from surface, d = 4R/3 - R = R/3 .
---------------
Attachments:
Answered by
88
Answer:R/3
Explanation:
Hi
See we know that potentialVc at center is always 3V/2
So Vc=3V/2
V' given is Vc/2
So V'=3V/4
That means that r=4R/3
So 4R/3 - R = R/3
Hope u get it
Click on the thank button and make it brainliest
Similar questions