A charge, q1 = 5.00uC, is at the origin, and a second charge, q2 = -3.00uC, is on the x-axis 0.800 m from the origin. Find the electric field at a point on the y-axis 0.500 m from the origin.
How do you solve this??? TT
Please show me the process.....
thanks!
:)
How do you solve this??? TT
Please show me the process.....
thanks!
:)
-
As you know, these are vector quantities.
E=kq/r^2
The E field from q1 at (0,0.5) is 9e9*5e-6/0.5^2 = 1.8e5 in the positive y direction
The E field from q2 at (0.8,0) is 9e9*(-3e-6)/(0.8^2+0.5^2) = -3.03e4 but has both horizontal and vertical components. The angle of the second field is arctan 0.8/0.5 = -58deg
Sum the vertical components: 1.8e5+3.03e4sin(-58) = 1.54e5 in the +y direction which makes sense because the positive charge is closer to 0,0.5 and is 5uC which is larger than 3uC
The horizontal component from q2 is 3.03e4cos(-58) = 1.61e4
The resultant is 1.55e5 at 84.1 degrees
pressed for time right now. Probably need to double check this later
E=kq/r^2
The E field from q1 at (0,0.5) is 9e9*5e-6/0.5^2 = 1.8e5 in the positive y direction
The E field from q2 at (0.8,0) is 9e9*(-3e-6)/(0.8^2+0.5^2) = -3.03e4 but has both horizontal and vertical components. The angle of the second field is arctan 0.8/0.5 = -58deg
Sum the vertical components: 1.8e5+3.03e4sin(-58) = 1.54e5 in the +y direction which makes sense because the positive charge is closer to 0,0.5 and is 5uC which is larger than 3uC
The horizontal component from q2 is 3.03e4cos(-58) = 1.61e4
The resultant is 1.55e5 at 84.1 degrees
pressed for time right now. Probably need to double check this later