The electric field between the plates of a parallel-plate capacitor is horizonta
ID: 2120717 • Letter: T
Question
The electric field between the plates of a parallel-plate capacitor is horizontal, uniform, and has a magnitude E=1.46*10^4 N/C. A small object of mass 0.0250 kg and charge -3.10 uC is suspended by a thread between the plates, as shown in the sketch. The tension in the thread is 0.249 N and the thread makes an angle of 10.5 degrees with the vertical.
Suppose the magnitude of the electric field is adjusted to give a tension of 0.260 N in the thread. This will also change the angle the thread makes with the vertical.
A.Find the new value of E. (N/C)
B.Find the new angle between the thread and the vertical. (degrees)
Explanation / Answer
FROM THE FIGUARE.....
B) T COS(THETA) = M*g.........(1)...
T SIN(THETA)M = q*E.....(2).........
COS(THETA) = M*g/T.......=(0.025*9.8)/(0.26).....=0.942
THETA=19.55 WITH THW VERTICAL................
A)E=T SIN(THETA)/q.....
E =0.26*SIN(19.55)/(-3.1*10^-6)...
E= 2.806*10^4 N/C
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.