When the top right-hand capacitor is filled with a material of dielectric constant κ, the charge on this capacitor is increases
by a factor of 1.35. Find the dielectric constant κ of the material inserted into the top right-hand capacitor.
2 answers:
Answer:
Explanation:
Given:
Capacitance, c1 = c2 = c3 = c4
= 80 micro farad
Voltage, v = Eb
= 4 V
We are not concerned with c3 and c4 just c1 and c2,
Since c1 and c2 are in series,
1/ct = 1/c1 + 1/c2
vab = v1 + v2
= q'/c1 + q'/c2
But c2 = κ × c,
= q'/c + q'/κc .....1
Let c1 = c2 = c,
Vab = q/c + q/c
= 2 × q/c. ........2
Equating equation 1 and 2,
q'/c + q'/κc = 2q/c
q'/c × (1 + 1/κ) = 2q/c
q'/q = 2κ/(1 + κ)
But q'/q = 1.35
2κ/(1 + κ) = 1.35
0.65κ = 1.35
κ = 1.35/0.65
= 2.077
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Answer:
a) 10.51 J
b) 3.48 m/s
Explanation:
Given data :
mass of train ( M ) = 2.2 kg
Given initial velocity ( u ) = 1.6 m/s
<u>a) calculating work done by the force over the journey of the train</u>
F = mx + b ------ ( 1 )
m = slope = ( Δ f / Δ x ) = 2.8 / -7.5 = - 0.373 N/m
x = distance travelled on the x axis by the train = 7.5 m
F = force experienced by the train = 2.8 N
x = 0
∴ b = 2.8
hence equation 1 can be written as
F = ( -0.373) x + 2.8 ----- ( 2 )
hence to determine the work done by the force
W = Note: the limits are actually 7.5 and 0
∴ W ( work done ) = -10.49 + 21 = 10.51 J
<u>b) calculate the speed of the train at the end of its journey</u>
we will apply the work energy theorem
W = 1/2 m*v^2 - 1/2 m*u^2
∴ V^2 = 2 / M ( W + 1/2 M*u^2 ) ( input values into equation )
V^2 = 12.11
hence V = 3.48 m/s