PH-2020-41 A negative charge, qi 10nc. is located at the origin of a coordinate
ID: 1318400 • Letter: P
Question
PH-2020-41 A negative charge, qi 10nc. is located at the origin of a coordinate system, with a Name: positive charge, qu +15nc, on the x-axis at x 5cm. picture showing the What is the net electric field on the y axis at y 12cm? Draw a clear notation. individual field vector well as the net field, Give your answer in unit vector b) For the same system of charges in part a) above, if an electron is placed at point y 12cm, what is the electric force on the electron? (Give both magnitude and direction) c) What is the electric potential (voltage aty 12cm due to the two charges?Explanation / Answer
We know that electric field is defined as = kq/r2 where k is constant , q is charge and r is distance between the charge and point.
(a) E1 = 6250 N/C , E2 = 7988.16 N/C
Now E2 is at angle therefore resolve it along x and y axis . and we see that the y component of E2 is in opposite direction to that of E1 therfore it will be subtracted and then by applying the pythagoras theorem we will calculate the electric field.
IN X direction = E2 Cos(67.38) = 3072.38 N/C (angle can be found by the right angle triangle property)
IN Y direction = E2Sin(67.38) - E1 = 1123.67 N/C
Now apply pythagoras theorm and find net magnitude
= 3271.41 N/C
Now for the direction Ey/Ex = tan(angle)
angle = 20.089 from -x direction.
(b) If an electron is placed at the point then we will apply the coloumb law.
As we have calcualted the electric field at that point so simply we have to multiply it with the charge of electron.
so it will be = 5234*10-19 N
(c) Electric potential = kq /r
we will calcualte the value of potential due to charge 1 and charge 2 and then finally we simply add them because it is a scalar quantity.
Potential due to charge 1 = 1038.46 V
Potential due to charge 2 = 750
therefore net potential = 1788.46 V
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