(a) Place a charge of -3.60 µC at point P and find the magnitude and direction o
ID: 1707296 • Letter: #
Question
(a) Place a charge of -3.60 µC at point P and find the magnitude and direction of the electric field at the location of q2.
E = N/C
= ° counterclockwise from the positive x-axis
(b) Find the magnitude and direction of the force on q2.
F = N
= ° counterclockwise from the positive x-axis
I Don't know how to do this Please help Me. Thank you. The top part is how i figured out the rest i think its the missing information to solve the bottom part.
E1 = 3.93 105 N/C Vector 1 is vertical, making an angle of 90° with respect to the positive x-axis. Use this fact to find its components. E1x = E1 cos(90°) = 0
E1y = E1 sin(90°) = 3.93 105 N/C Next, find the magnitude of 2, again with Equation 15.6.
E2 = 1.92 105 N/C Obtain the x-component of 2, using the triangle in Figure 15.12 to find cos .
E2x = E2 cos = (1.92 105 N/C)(0.600)
E2x = 1.15 105 N/C Obtain the y-component in the same way, but a minus sign has to be provided for sin because this component is directed downward.
E2y = E2 sin = (1.92 105 N/C)(0.800)
E2y = -1.54 105 N/C Sum the x-components to get the x-component of the resultant vector. Ex = E1x2x = 0 + 1.15 105 N/C
Ex = 1 N/C
+ E Sum the y-components to get the y-component of the resultant vector. Ey = E1y2y = 3.93 105 N/C - 1.54 105 N/C
Ey = 2 N/C
+ E Use the Pythagorean theorem to find the magnitude of the resultant vector.
E = 3 N/C
The inverse tangent function yields the direction of the resultant vector.
= 4°
(b) Find the force on a charge of 2.00 10-8 C placed at P. Calculate the magnitude of the force (the direction is the same as that of because the charge is positive). F = Eq = E(2.00 10-8 C)
F = 5 N