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  1. G

    Modulus of the electric field created by a sphere

    I think the right solution is c). I'll pass on my reasoning to you: R=6\, \textrm{cm}=0'06\, \textrm{m} \sigma =\dfrac{10}{\pi} \, \textrm{nC/m}^2=\dfrac{1\cdot 10^{-8}}{\pi}\, \textrm{C/m}^2 P=0'03\, \textrm{m} P'=10\, \textrm{cm}=0,1\, \textrm{m} Point P: \left. \phi =\oint E\cdot...
  2. G

    Origin of potentials

    I think the right choice is c. I'll pass on my reasoning to you: We can think that if the formula of the potential is V(r)=\dfrac{kq}{r} If r tends to infinity, then V(r)=0. But the correct answer is d).
  3. G

    Potential at a point

    I thought the right choice was d). But when it comes to the solutions, it is b) and I don't understand why. My reasoning would be: the potential at a point is the work that the electric field does to transport a charge from infinity to that point, so if the field is zero, it does no work and...
  4. G

    Capacitance of a spherical capacitor

    I think I have the solution. Is that right? \left. \phi =\oint \vec{E}\cdot d\vec{S}=\oint E\cdot dS\cdot \underbrace{\cos 0}_1=E\oint dS=E\cdot S \atop \phi =\dfrac{Q_{enc}}{\varepsilon_0 \varepsilon_r}=\dfrac{Q}{\varepsilon_0 \varepsilon_r}=\dfrac{\sigma \cdot S}{\varepsilon_0 \varepsilon_r}...
  5. G

    Capacitance of a spherical capacitor

    When I try to do Gauss, the permeability is not always that of the free space, but it varies: up to a certain radius it is that of the void and then it is the relative one. How can I relate them? I'm trying to calculate the capacity of a spherical capacitor. The scheme looks like this: inside I...
  6. G

    Electric field and electric potential exercise

    Okay, perfect. Gauss law would say that the charge inside a closed surface (which can take any shape) is proportional to the flux through it, wouldn't it?
  7. G

    A rocket and its gas exhaust velocity

    Right. Well, it must be mistaken.
  8. G

    A rocket and its gas exhaust velocity

    "I) Transfer from initial (circular) to transfer (elliptical) orbit. - The initial trajectory of m is a circumference centred in the massive object M of radius r=r_p . The velocity of m is constant, and can be easily obtained as follows: e=-\dfrac{GM}{2r_p}=\dfrac12 v_p^2-\dfrac{GM}{r_p}...
  9. G

    A rocket and its gas exhaust velocity

    And here you substitute the velocity for the expression of u\cdot \ln \left( \dfrac{m_o}{m_f}\right) and isolate the mass
  10. G

    A rocket and its gas exhaust velocity

    I also think that, but I think it might indicate the speed variation. Otherwise, there's no meaning in the minus sign. If you do the velocity variation considering circular trajectories I think that's correct. Since the speed: V=\sqrt{\dfrac{GM}{R}} Although not on top it shouldn't have speed
  11. G

    A rocket and its gas exhaust velocity

    Yes, in this problem it is considered to burn all the fuel instantly at first and continue until gravity cancels out all the initial speed.
  12. G

    Electric field and electric potential exercise

    Aa okay, that is even if it is not constant you know that it will surely be proportional because it must always be a multiple of the elementary charge, the electron. Right?
  13. G

    A rocket and its gas exhaust velocity

    You can find the speed with \Delta V=u\cdot \ln \left( \dfrac{m_0}{m_f}\right) It's Tsiolkovski's equation.
  14. G

    A rocket and its gas exhaust velocity

    Don't worry about the rocket part, it's just this step
  15. G

    A rocket and its gas exhaust velocity

    This isn't right, is it? -\dfrac{GM}{R}+\dfrac12 v^2=-\dfrac{GM}{R+h} v=\sqrt{\dfrac{GM}{R}}\left( 1-\sqrt{\dfrac{R}{R+h}}\right) He's doing energy conservation. The mechanical energy at the Earth's surface is equal to the energy when the speed is 0.
  16. G

    Electric field and electric potential exercise

    a) \vec{F}=\vec{E}\cdot q \phi =\oint \vec{E}d\vec{S}=\oint \vec{E}d\vec{S}=\underbrace{\oint \vec{E}d\vec{S}}_{\textrm{FACES } \perp}+\underbrace{\oint \vec{E}d\vec{S}}_{\textrm{FACES } \parallel}=0+\oint EdS\cdot \underbrace{\cos 0}_1= E2S \dfrac{Q_{enc}}{\varepsilon_0}=\phi \left...
  17. G

    Determing the center of gravity of a shaded section

    ##\bar{x}=\dfrac{\frac{7R}{3}\cdot R^2+R\cdot 4R^2-\frac{4R}{3\pi}\cdot \frac{\pi R^2}{4}}{\frac{20R^2-\pi R^2}{4}}=\dfrac{\frac{7R^3}{3}+4R^3-\frac{4\pi R^3}{12\pi}}{\frac{R^2\cdot (20-\pi)}{4}}=\dfrac{\frac{7R^3}{3}+4R^3-\frac{R^3}{3}}{\frac{R^2\cdot...
  18. G

    Determing the center of gravity of a shaded section

    Writing in terms of ##R##: a) ##A_T=2R\cdot 2R-\dfrac14 \pi R^2+\dfrac{R2R}{2}=4R^2-\dfrac{\pi R^2}{4}+R^2=\dfrac{16R^2-\pi R^2+4R^2}{4}=\dfrac{R^2\cdot (20-\pi)}{4}=\dfrac{20R^2-\pi R^2}{4}=143,3\, \textrm{cm}=1,433\, \textrm{m}## b) Centers of gravity: - Figure 1...
  19. G

    Determing the center of gravity of a shaded section

    Okay, I'll do it by leaving R indicated and I'll pass it around. You're absolutely right.
  20. G

    Determing the center of gravity of a shaded section

    Determine the volume of the shaded area around the Y-axis by using the theorem of Pappus Guldinus, where value of R = 143,3 cm. a) Determine the area of the shaded section. b) Determine the center of gravity of the shaded section. c) Detrmine the volume by using the theorem of Pappus Guldinus...
  21. G

    Determining a centroid

    Yes, I applied this second method in the end, i.e. integrating with respect to ##x## and considering that ##y## was in the middle of the differential.
  22. G

    Determining a centroid

    Okay perfect thank you!
  23. G

    Determining a centroid

    Summary:: I'm solving an exercise. I have the following center of gravity problem: Having the function Y(x)=96,4*x(100-x) cm, where X is the horizontal axis and Y is the vertical axis, ranged between the interval (0, 93,7) cm. Determine: a) Area bounded by this function, axis X and the line...
  24. G

    Gauss-Theorem on a solid dielectric sphere

    The load system formed by the point load and the load distribution generates two regions in space corresponding to r<1m and r>1m, i.e. inside and outside the sphere. Given the symmetry of the distribution, by means of the Gaussian theorem we can find the modulus of the field at a distance r from...
  25. G

    Electric dipoles

    Clear!! And is it hard to know why it has that sense?
  26. G

    Electric dipoles

    I found this: An electric dipole consists of two charges, one positive +q and the other negative -q of the same value, separated by a distance d, usually small. The main characteristic of the electric dipole is the dipole moment, which is defined as the product of the charge q by the distance...
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