Part A
1 answer:
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Answer:
a) E = 4.5*10⁴ V/m
b) C= 17.7 nF
c) Q = 159. 3 nC
Explanation:
a)
- By definition, the electric field is the electrostatic force per unit charge, and since the potential difference between plates is just the work done by the field, divided by the charge, assuming a uniform electric field, if V is the potential difference between plates, and d is the separation between plates, the electric field can be expressed as follows:
b)
- For a parallel-plate capacitor, applying the definition of capacitance as the quotient between the charge on one of the plates and the potential difference between them, and assuming a uniform surface charge density σ, we get:
From (1), we know that V = E*d, but at the same time, applying Gauss'
Law at a closed surface half within the plate, half outside it , it can be
showed than E= σ/ε₀, so finally we get:
c)
- From (3) we can solve for Q as follows:
A. 0.77 A
Using the relationship:
where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.
For the first light bulb, P = 60 W and V = 120 V, so the resistance is
For the second light bulb, P = 200 W and V = 120 V, so the resistance is
The two light bulbs are connected in series, so their equivalent resistance is
The two light bulbs are connected to a voltage of
V = 240 V
So we can find the current through the two bulbs by using Ohm's law:
B. 142.3 W
The power dissipated in the first bulb is given by:
where
I = 0.77 A is the current
is the resistance of the bulb
Substituting numbers, we get
C. 42.7 W
The power dissipated in the second bulb is given by:
where
I = 0.77 A is the current
is the resistance of the bulb
Substituting numbers, we get
D. The 60-W bulb burns out very quickly
The power dissipated by the resistance of each light bulb is equal to:
where
E is the amount of energy dissipated
t is the time interval
From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.