(a) Find the magnitude and direction of the electric field at x = -1 m, y = 0 .
ID: 2124642 • Letter: #
Question
(a) Find the magnitude and direction of the electric field at x = -1 m, y = 0.
___ N/C
___ degrees (counterclockwise from plus x axis)
(b) Calculate the magnitude and direction of the force on an electron at x = -1 m, y = 0 .
___ N
___ degrees counterclockwise from plus x axis
A point charge of -7 mu C is located at x = 2 m, y = -2 m. A second point charge of 12 mu C is located at x = 1 m, y = 3 m. Find the magnitude and direction of the electric field at x = -1 m, y = 0. Calculate the magnitude and direction of the force on an electron at x = -1 m, y = 0 .Explanation / Answer
a.Let charge q1: 4i - 2j -3 uC
Let charge q2: i + 2j 12 uC
Let point P: -i + 0j
The field at a point due to a point charge Q is:E = KQ r cap /r^2 that's the vector form of it:
The field due to charge q1 is:E1 = Kq1/-1-4)^2 + 0+2)^2 = -5i+2j/sqrt(29)
The field due to q2 is:-9535i -9535 jN/C
Add the fields together to get the total field:
The magnitude is given by:E = 7690 N/C
At an angle of:theta = 282.14 deg 0r -77.86 deg from +x axis.
The force on a point charge is: F = Eq
The electron has a charge of: -1.60*10^-19
The magnitude of the force it experiences is:
F =|E||q| = 1.23*10^-15 N
At angle which is 180 degrees from the field due to the negative charge of the electron, i.e.
? = 49 degrees
c. Because the electron has a negative charge, the angle will be rotated by 180 degrees.
You could add or subtract 180, but since you want the angle to be between:
0 and 360, or -180 and 180, you should just subtract 180.
282.140 - 180 = 102.14 degrees.
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