A student writes: “When I rub my dry hands together on a cold morning, they warm up. However, when I repeat this with soapy wate
2 answers:
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Answer:
30 cm
Explanation:
To solve this problem, we use the lens equation:
where
f is the focal length of the lens
p is the distance of the object from the lens
q is the distance of the image from the lens
In this problem, we know that
q = -30 cm is the distance of the image from the lens; it is negative because the image is formed in front of the lens, so it is a virtual image
We also know that the size of the image is twice that of the object, so the magnification is 2:
where M is the magnification of the lens. Solving this equation for p,
So, the distance of the object from the lens is 15 cm.
Now we can finally solve the lens equation to find f, the focal length:
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: