Новое решение
en/6.3.7.md
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| + | ### Statement | ||
| + | |||
| + | $6.3.7.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | As the statement of the problem doesn't clarify if the sphere is conducting or non-conductive, let's consider both cases. | ||
| + | Case 1) Nin-Conductive sphere: | ||
| + | Let's assume that charge is uniformly distributed in the volume. Let electrical potential be: | ||
| + | $V(r) = -\int_{\infty}^{r} \vec{E} \cdot d\vecP{r}$ (1) | ||
| + | for the center, let's consider $r<<R$, or $r \rightarrow 0$. | ||
| + | Applying Gauss Law: | ||
| + | For $r<R$, | ||
| + | $\int \vec{E} \cdot d\vec{S} = \frac{q_{enc}}{\varepsilon_0}$ | ||
| + | but the enclosed charge into a sphere of radius r is related to the charge distribution per unit of volume, | ||
| + | $\rho = \frac{Q}{\frac{4}{3}\pi R^3} = \frac{q_{enc}}{\frac{4}{3}\pi r^3}$ | ||
| + | $q_{enc} = Q\left(\frac{r}{R}\right)^3$ | ||
| + | so, | ||
| + | $E(r) = \frac{Q r}{4\pi\varepsilon_0 R^3}\;\;\;\;\forall\;r < R$ | ||
| + | and for $r > R$, the enclosed charge is Q, then, | ||
| + | $E(r) = \frac{Q}{4\pi\varepsilon_0 r^2}\;\;\;\;\forall\;r>R$ (2) | ||
| + | This mean that function $E(r)$ has two behaviors, depending on values of $r$. According (1) and assuming $r\rightarrow 0$, | ||
| + | $V(r) = -\left(\int_{\infty}^{R} \vec{E} \cdot d\vec{r} + \int_{R}^{r} \vec{E} \cdot d\vec{r}$ (I) | ||
| + | developing, | ||
| + | $V(r) = \frac{Q}{4\pi\varepsilon_0 R} - \frac{Q}{8\pi\varepsilon_0 R^3}(r^2-R^2)$ | ||
| + | as $r$ tends to zero, | ||
| + | $V(r) = \frac{3Q}{8\pi\varepsilon_0 R^3}$ (3) | ||
| + | Case 2) Conducting sphere | ||
| + | In this case, charge stays on sphere surface, so the electric field inside of it is null ($E(r) = 0$), so potential is constant inside the sphere and coincides with the value of it on the surface. According to (1) and (2), | ||
| + | $V(r) = -\int_{\inty}^{R} \frac{Q}{4\pi\varepsilon_0 r^2}dr$ | ||
| + | $V(0) = V(r) = \frac{Q}{4\pi\varepsilon_0 R}$ (4) | ||
| + | Finally, if sphere is non-conductive $V(0)$ depends on the charge distribution, as we saw in (I), but in the second case (maybe the problem refers specifically to a conducting sphere), it \textbf{doesn't depend} on it. If the charge is non-uniformly distributed, the potential on the surface \textbf{does change} with the local distribution. For example, charge accumulations in certain areas generate angular variations in the potential. However, outside the sphere, at a great distance, the potential depends only on the total charge, Q, as if it were a point charge at the center. | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $6.3.7.$ [Insert the problem statement] | |||
| ### Solution | |||
| As the statement of the problem doesn't clarify if the sphere is conducting or non-conductive, let's consider both cases. | |||
| Case 1) Nin-Conductive sphere: | |||
| Let's assume that charge is uniformly distributed in the volume. Let electrical potential be: | |||
| $V(r) = -\int_{\infty}^{r} \vec{E} \cdot d\vecP{r}$ (1) | |||
| for the center, let's consider $r<<R$, or $r \rightarrow 0$. | |||
| Applying Gauss Law: | |||
| For $r<R$, | |||
| $\int \vec{E} \cdot d\vec{S} = \frac{q_{enc}}{\varepsilon_0}$ | |||
| but the enclosed charge into a sphere of radius r is related to the charge distribution per unit of volume, | |||
| $\rho = \frac{Q}{\frac{4}{3}\pi R^3} = \frac{q_{enc}}{\frac{4}{3}\pi r^3}$ | |||
| $q_{enc} = Q\left(\frac{r}{R}\right)^3$ | |||
| so, | |||
| $E(r) = \frac{Q r}{4\pi\varepsilon_0 R^3}\;\;\;\;\forall\;r < R$ | |||
| and for $r > R$, the enclosed charge is Q, then, | |||
| $E(r) = \frac{Q}{4\pi\varepsilon_0 r^2}\;\;\;\;\forall\;r>R$ (2) | |||
| This mean that function $E(r)$ has two behaviors, depending on values of $r$. According (1) and assuming $r\rightarrow 0$, | |||
| $V(r) = -\left(\int_{\infty}^{R} \vec{E} \cdot d\vec{r} + \int_{R}^{r} \vec{E} \cdot d\vec{r}$ (I) | |||
| developing, | |||
| $V(r) = \frac{Q}{4\pi\varepsilon_0 R} - \frac{Q}{8\pi\varepsilon_0 R^3}(r^2-R^2)$ | |||
| as $r$ tends to zero, | |||
| $V(r) = \frac{3Q}{8\pi\varepsilon_0 R^3}$ (3) | |||
| Case 2) Conducting sphere | |||
| In this case, charge stays on sphere surface, so the electric field inside of it is null ($E(r) = 0$), so potential is constant inside the sphere and coincides with the value of it on the surface. According to (1) and (2), | |||
| $V(r) = -\int_{\inty}^{R} \frac{Q}{4\pi\varepsilon_0 r^2}dr$ | |||
| $V(0) = V(r) = \frac{Q}{4\pi\varepsilon_0 R}$ (4) | |||
| Finally, if sphere is non-conductive $V(0)$ depends on the charge distribution, as we saw in (I), but in the second case (maybe the problem refers specifically to a conducting sphere), it \textbf{doesn't depend} on it. If the charge is non-uniformly distributed, the potential on the surface \textbf{does change} with the local distribution. For example, charge accumulations in certain areas generate angular variations in the potential. However, outside the sphere, at a great distance, the potential depends only on the total charge, Q, as if it were a point charge at the center. | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||