A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 44.1m/s^2 . The acce
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Answer:
the resistance of the wire has no effect on the brightness of the bulb.
Explanation:
Let's apply ohm's law for your light bulb circuit plus wires plus switch
V = I R_{bulb} + I R_ {wire}
the current in a series circuit is constant
V = I (R_{bulb} + R_{wire})
To know the effect of the wires on the brightness of the bulb, we must look for the value of the typical resistance of these elements.
Incandescent bulb
Power 60 W
let's use the power ratio
P = V I = V2 / R
R = V2 / P
the voltage value for this power is V = 120 V
R = 120 2/60
R_bulb = 240 Ω
Resistance of a 14 gauge copper wire (most used), we look for it on the internet
R = 8.45 Ω/ km
in a laboratory circuit approximately 2 m is used, so the resistance of our cable is
R = 8.45 10⁻³ 2
R_wire = 0.0169 Ω
let's buy the two resistors
R_{bulb} = 240
R_{wire} = 0.0169
therefore resistance of the bulb is much greater than that of the wire, therefore almost all the power is dissipated in the bulb.
In summary, the resistance of the wire has no effect on the brightness of the bulb.
Answer:
a) 4.33 pC b) 5.44*10² N/C
Explanation:
a) The vertical deflecting plates of an oscilloscope form a parallel-plate capacitor.
The value of the capacitance, for a parallel-plate capacitor with air dielectric, can be found to be as follows, applying Gauss' law to the surface of one of the plates, and assuming a uniform surface charge density:
C = ε₀*A / d
where ε₀ = 8.85*10⁻¹² F/m, A = (0.03m)², and d = 0.046 m (we assume that the informed value of 4.6 m is a typo, as no oscilloscope exists with this separation between plates).
Replacing by these values, we find the equivalent capacitance of the plates, as follows:
By definition, the capacitance of any capacitor can be expressed as follows:
where Q= charge on any of the plates, and V= potential difference between them.
As we know C and V, we can find Q as follows:
Q = C*V = 0.173*10⁻¹² F * 25.0 V = 4.33*10⁻¹² C = 4.33 pC
b) We can find the electric field in several ways, but one very easy is applying Gauss' Law to a pillbox with a face outside one of the plates (paralllel to it) and the other inside the surface.
The total electric flux through the surface must be equal to the enclosed charge, divided by ε₀.
If we look to the flux crossin any face, we find that the only one that has a non-zero flux, is the one outside the surface.
As the electric crossing the boundary must be normal to the surface (in electrostatic conditions, no tangential field can exist on the surface) , and we assume that the surface charge density that creates it is constant across the surface, we can write the Gauss ' Law as follows:
E*A = Q / ε₀
where A = area of the plate = (.03m)² = 9*10⁻⁴ m², Q= charge on one of the plates = 4.33*10⁻¹² C (as we found in a)) and ε₀ = 8.85*10⁻¹² N/C.
Replacing by these values, and solving for E, we have:
⇒ E = 5.44*10² N/C