(a) what is the magnitude of the force on the charge?

Question

(a) cont... A charge of q=+7.65 micro coulomb is located in an electric field. The x & y components of the electric field are Ex= 6.41x10+3 N/C & Ey= 8.33x10+3 N/C, respectively.
Also determine the angle that the force makes with the +x axis

(b) An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is 1.63x10-7 N/m^2, & the plates are separated by a distance of 1.44x10-2 m. How fast is the electron moving just before it reaches the positive plate?

(c) A small plastic ball of mass 6.20x10-3 kg & charge +0.19 micro coulomb is suspended from an insulating thread & hangs between the plates of a capacitor. The ball is in equilibrium, with the thread making an angle of 30º with respect to the vertical. The area of each plate is 0.0128m^2. What is the magnitude of the charge on each plate?

Answer 1

a) F=qE
Fx=q Ex
Fy= q Ey

b)Hmmm... "...charge per unit area on each plate is 1.63x10-7 N/m^2" Interesting

c)
F= F1 + F2
F1= k q Q1/R^2
F1= k q Q2/R^2
Assuming that Q1=Q2=Q but apposite in sign

F= 2kq Q/R^2
R= .5 D d- distance between the plates
also
F= mg sin(A)

Then Q= mg R^2 sin(A)/(2k q)

Answer 2

so as which you have have been given 2 forces. each and every has a value of F. the perspective between them is a hundred twenty ranges. So use the regulation of cosines to get the consequent tension. c^2 = a^2 + b^2 + 2ab cos a hundred twenty I placed plus fairly of minus as a results of fact the sole hundred twenty degree attitude is a hundred and 80 - the perspective contained interior the triangle. a=b=F c^2 = F^2 + F^2 + 2F^2 cos a hundred twenty = F^2 so c = F (B) think of roughly guessed this intuitively. a nil.33 tension of equivalent value placed on the 0.33 vertex could make the sum 0 by technique of symmetry. hence, the cost of the sum of two must be reminiscent of the cost of one.