Answer:
height above the fixed charge is 8.571 cm
Explanation:
given data
mass m = 0.050 g = 5 × 
kg
charge bead q1 = 20 nC = 20 × 
C
charge base q2 = 20 nC = 20× C
to find out
what height above the fixed charge does the bead rest
solution
we know that when charge at rest then downward gravitational force is balance by electrostatic force so
.............1
here k is 9 × 
Nm²/C² and g = 9.8 m/s² and r is height of bread
put here all value in equation 1
r² = 7.3469 × 
r = 0.08571 m = 8.571 cm
so height above the fixed charge is 8.571 cm
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)
The ball took half of the total time ... 4 seconds ... to reach its highest
point, where it began to fall back down to the point of release.
At its highest point, its velocity changed from upward to downward.
At that instant, its velocity was zero.
The acceleration of gravity is 9.8 m/s². That means that an object that's
acted on only by gravity gains 9.8 m/s of downward speed every second.
-- If the object is falling downward, it moves 9.8 m/s faster every second.
-- If the object is tossed upward, it moves 9.8 m/s slower every second.
The ball took 4 seconds to lose all of its upward speed. So it must have
been thrown upward at (4 x 9.8 m/s) = 39.2 m/s .
(That's about 87.7 mph straight up. Somebody had an amazing pitching arm.)
Answer:
The increase in potential energy of the ball is 115.82 J
Explanation:
Conceptual analysis
Potential Energy (U) is the energy of a body located at a certain height (h) above the ground and is calculated as follows:
U = m × g × h
U: Potential Energy in Joules (J)
m: mass in kg
g: acceleration due to gravity in m/s²
h: height in m
Equivalences
1 kg = 1000 g
1 ft = 0.3048 m
1 N = 1 (kg×m)/s²
1 J = N × m
Known data
Problem development
ΔU: Potential energy change
ΔU = U₂ - U₁
U₂ - U₁ = mₓgₓh₂ - mₓgₓh₁
U₂ - U₁ = mₓg(h₂ - h₁)
The increase in potential energy of the ball is 115.82 J
