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:
Explanation:
(Assuming the cell in the circuit has 0 internal resistance)
Ohm's Law is given as:
Voltage is Current multiplied by Resistance.
We can rearrange this formula to give us:
Now we can plug in our values