00:01
For this problem on the topic of electrostatics, for the given figure we have charged q1 to be 1 times 10 to the minus 7 kuloms and charge q2 2 2 times 10 to the minus 7 couloms.
00:11
We want to find the electric field at the point 0 and 3 centimeters and then the force that will act on an electron at the at that position.
00:21
So firstly we want to find the x component of the field e x and this is the x component of the field due to the first charge e1 x.
00:31
Plus the x component of the field due to the second charge e2x.
00:36
And so we know this is kq1 over r1 squared times the cosine of theta 1 plus k q2 over r2 squared cosine theta 2 and we can write this as k into q1 over r1 squared cosine theta 1 plus q2 over r2 squared cosine theta 2.
01:14
Putting in our values we get this to be 8 .99 times 10 to the 9 newton meter squared per coulom squared into 1 times 10 to the 10 to the minus 7 couloms divided by 0.
01:42
0 .05 meters squared plus 0 .03 meters squared, multiplied by the cosine of theta 1, which is 0 .05 meters divided by the square root of 0 .05 meters squared plus 0 .05 meters squared plus 0 .0 .0 5 meters squared plus 0 .0 .0.
02:14
3 meters squared plus the second term, which is 2 times 10 to the minus 7 columns divided by 0 .0 8 meters squared plus 0 .03 meters squared into cosine of theta 2, which is 0 .0 .0 .0 .0 3 meters squared into cosine of theta 2, which is 0 .0 .0 .0...