Here are the steps you would need to follow:
#1). Define what 'the 'X is, and how it's related to the ball.
#2). Be clear on how 'the X' is related to the 'known velocity'.
#3). Identify how the 'known velocity' is related to the action of the ball when it's launched.
With this information in front of you, you'll have a much better chance
of answering the question.
With none of it in front of me, I have no chance at all.
Answer:
(a) Since net charge remains same,after immersion Q is same
(b) I. 14.56pF ii. 3.05V
(c) ΔU = 5.204nJ
Explanation:
a)
C = kεA/d
k=1 for air
ε is 8.85x10-12F/m
A = .0025m2
d = .125m
C = 8.85x10-12x.0025/.125 = 1.77x10-13F = 0.177pF
Q = CV = .177pF * 244V = 43.188pC
Since net charge remains same,after immersion Q is same
b)
C = kεA/d, for distilled water k is approx. 80
Cwater = Cair x k
=0.177pF x 80 = 14.16pF
Q is same and C is changed V=Q/c holds. where Q is still 43.188pC and C is now 14.16pF, so V = 43.188pC/14.16pF = 3.05V
c) Change in energy: ΔU = Uwater - Uair
Uwater = Q2/2C = (43.188)2/2x.177pF = 5.27nJ
Uair = Q2/2C = (43.188)2/2x14.16pF = 0.066nJ
ΔU = 5.204nJ
<h3>NET FORCE </h3>
The net force is equal to the sum of two forces acting in the same direction on an object. The net force is always bigger than the individual forces in this scenario.
