00:01
Hi students in this question a uniformly charged sphere is been given and in the first part of the question we are asked to find out the volume charge density row and the equation of row is taken as row to be equal to charge q divided by volume v let this be equation number one and here this charge queue can be found out from the equation of electric field which is taken as electric field e to be equal to coulom's constant k into charged q divided by r square where this r corresponds to this from the center of the sphere.
00:31
Now here a few data are been given which include the value of e which is taken to be 990 newton per kulum.
00:39
The value of r is taken to be 8 into 10 raise to minus 2 meter and finally the coulomb's constant k is taken to be 9 into 10 raise to 9 newton meter square per column square.
00:52
Now by substituting these three values in the above equation, our equation changes as 990 to be equal to 9.
01:00
9.
01:00
10 raised to 9 into the value of q divided by 8 into 10 raise to minus 2 the whole square which on simplification gives the value of charge q to be equal to 7 .04 into 10 raised to minus 10 kool -up.
01:15
Let this be equation number 2.
01:17
Next we have to find out the volume of the sphere which can be found out using the equation volume v to be equal to 4 by 3 into 5 r cube.
01:27
And here the value of radius of the sphere r is taken to be a value of 4 into 10 raise to minus 2 meter.
01:35
Now by substituting this value of radius r in the above equation, we get volume v to be equal to 4 by 3 into 3 .14 into 4 into 10 raise to minus 2 the whole cube...