1.Four point charges are fixed at the corners of a square of side a=0.1m. each c
ID: 2257275 • Letter: 1
Question
1.Four point charges are fixed at the corners of a square of side a=0.1m. each charge is q,q,-2q,2q. Charge q = 10.uC.
A determine the net electri field at the vector at point O, the center of the square. B) if a electron was placed at the center O what is its direction of motion.
2.Two identical non conducting spheres are held together along the xaxis. The radius R of each sphere is 0.1m. A charge odfQ=10.0uC is uniformly distributes throughout eachh sphere. Determine the net eletric field vector due to these charge distributioms at points
i)x=0m
ii)x=R/2
iii)x=Rm
iv)x=2Rm
B) What is the result of releasing the spheres
3)A suspect finds himself being chased by a trooper car and at the same time beinf approached from the opposite direction by a second trooper car.Both troopers are using identical sirens of frequency f=10kHz. The suspect distimctly hears a beat frequency of 583 Hz. Determine the speed at whih the suspect is running from and to the two respective troopers?
4) A set of six identical resisters R=11.0 ohms is configured as a circuit across a battery supplying 100.0V as shown.
a)what is R effecitive of the circuit
b)the total current supplied by the battery
c)the voltage across the resistors at A and B
d) the voltage across the branch D
Explanation / Answer
1 E = kQ*2^0.5/a^2 = 1272.9 - j
B ) positve y dircetion
2)
distance between sphere is required ... here I am doing assuming both spehre toughes each other.
X =0 , E = KQ/R^2 = 9000 N/c
X =R/2 , E = KQ/(2R^2) - KQ/(2.25R^2) = 500 N/C
X = R , E =0
X = 2R , E = -9000 N/C
3) F1 = 10 -0.58 = 9.42 KhZ
F2 = 10.58 KHZ
F2 = (V+340)*10/(340 -V1)
F1 = (340-V)*10/(340-V2)
4) Figure required , I am assuming all resitor are in parallel
i) Ref =11/6 =1.83
ii) i = current =100/1.83 =54.6 A
iii) VAB = 100
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.