00:01
Okay, so this is the equation for the electric view of the uniform ring with the charge, which is equal to k -e -x over a square plus x -square to the power of three over two times charge cube.
00:12
A is a radius, which is given as 10 .0 centimeter, which is 0 .1 meter.
00:18
And x is the distance from the center of ring.
00:21
And in question a, it's given as 1 .0 -0 centimeter, which is 0 .01 meter.
00:26
And k .e is a coolant constant, which is 8 .99 times 10 .2 .9, newtons.
00:31
Square per kolon square and chart q is given at 75 .0 microcourt which is 75 times 10 to 96cccccccccc so let's plug in all these values back into the equation and we have electric field is equal to oh sorry 8 .99 times 10 to power 9 newton times meter per coolon square and then times 0 .01 meter over a square which is uh 0 .1 meter square plus 0 .01 meter and square, then to the power of 2, 3 over 2.
01:25
Okay? and then has charged q, which is 75 times 10 to the power of negative 6 cologne.
01:36
And this will give us open.
01:40
The electric view for question a is about 6 .64 times 10 to the the power of 6 newton per cologne.
01:54
And for question b, we use the same equation.
01:58
By this time, the x, which is the distance from the center of ring, is given as 5 .0 centimeter.
02:04
If we convert meter, it's 0 .05 meter.
02:06
So let's try to determine the electric view for this case, which is e is equal to 8 .99 times 10 to the power of 9 newton per times meter square per colon square...