A point charge q1 = -5.7 μC is located at the center of a thick conducting spherical shell of inner radius a = 7.4 cm and outer
radius b = 11.5 cm. The conducting spherical shell has a net charge of q2 = 2.6 μC. A) Draw the electric field lines inside the inner radius, between the inner and outer radius, and exterior to the conducting shell. B) What is the magnitude of the electric field at a location 8.0 cm from the point charge? C) What is the electric field at the point θ = 45o , r = 13 cm?
1 answer:
Answer: a) see attach file; b) E=0; c)E=-1.65* 10^6 N/C
Explanation: In order to solve this problem we have to use the gaussian law which is given by:
∫E*dr=Q inside/εo
E for r=8 cm is located inside the conducting shell so E=0.
For r=13 cm we have to use above gaussian law considering the total charge inside a gaussian surface with radius equal to 0.13 m. ( see attach)
You might be interested in
Answer:
Explanation:
A. The earth about its axis:
The earth makes one revolution in 24 hours. to know the number of revolutions per second it makes, we need to convert hours to seconds and the revolution to rad.
B. The minute hand of the clock makes one revolution in 60 minutes
To convert this to rad per second, we have
C. The hour hand of a clock completes one revolution in 12 hours
D. an egg beater turning at 300 rpm.
Answer:
puck decelerates due to the kinetic frictional force μk mg
Explanation:
given data
total distance = 12 m
coefficient of kinetic friction = 0.28
solution
we will apply equation of motion that is
v² - u² = 2 × a × s ................1
we know acceleration will be
a =
Then we have
Force = mass × acceleration .................2
m × = -μk mg
The puck decelerates due to the kinetic frictional force μk mg
and frictional force is negative as it opposes the motion.
so we get initial velocity of the puck which is strike.