Edits to “Statement”, “Solution”, “Field at Point A”

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en/6.2.12.md +34 −52
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−### Statement
+### Statement
−\\\\\\\$6.2.12.\\\\\\\$ In a uniformly charged infinite plate, a spherical cavity was cut out as shown in the figure. Plate's thickness is \\\\\\\$h\\\\\\\$ and its bulk charge density is \\\\\\\$\\\\\\\\rho\\\\\\\$. What is the electric field strength at point \\\\\\\$A\\\\\\\$? At point \\\\\\\$B\\\\\\\$? Find the dependence of the electric field strength along the line \\\\\\\$OA\\\\\\\$ on the distance to the point \\\\\\\$O\\\\\\\$.
+**6.2.12.** In a uniformly charged infinite plate, a spherical cavity was cut out as shown in the figure. The plate's thickness is $h$ and its bulk charge density is $\rho$.
−![ For problem \\\\\\\$6.2.12\\\\\\\$ |509x300, 59%](../../img/6.2.12/statement.png)
+1. What is the electric field strength at point $A$ (the center of the cavity)?
+2. What is the electric field strength at point $B$ (on the surface of the plate directly above the center)?
+3. Find the dependence of the electric field strength along the line $OA$ on the distance $r$ from the center $O$.
+---
+
### Solution
−The idea for solving this problem is to consider superposition principle: as cavity is uncharged, we can suppose that it exists a superposition of a plate (with charge density \\\\\\\$\\\\\\\\rho\\\\\\\$) and a sphere with charge density \\\\\\\$-\\\\\\\\rho\\\\\\\$. So, if we analyze the field in \\\\\\\$A\\\\\\\$ without cavity, we have two plates of thickness \\\\\\\$\\\\\\\\frac{h}{2}\\\\\\\$, one over \\\\\\\$A\\\\\\\$ and the other one under A, both with charge density \\\\\\\$\\\\\\\\rho\\\\\\\$, then field in \\\\\\\$A\\\\\\\$ is null. While for \\\\\\\$A\\\\\\\$ in the sphere's surface, applying Gauss's Law,
+The key to solving this problem is the **principle of superposition**. We can treat the cavity as a combination of a solid, uniformly charged plate (density $+\rho$) and a sphere of the same size with an opposite charge density ($-\rho$).
−\\\\\\\$\\\\\\\$
−-E_A\\\\\\\\cdot 4\\\\\\\\pi\\\\\\\\left(\\\\\\\\frac{h}{2}\\\\\\\\right)^2 = -\\\\\\\\frac{q_{enc}}{\\\\\\\\varepsilon_0}
−\\\\\\\$\\\\\\\$
+#### Field at Point A
+In a solid infinite plate of thickness $h$, the electric field at the exact center (point $A$) is zero due to symmetry. Therefore, the field at $A$ is generated solely by the "negative" sphere.
−as \\\\\\\$V = \\\\\\\\frac{4}{3}\\\\\\\\pi\\\\\\\\left(\\\\\\\\frac{h}{2}\\\\\\\\right)^3\\\\\\\$, so \\\\\\\$6\\\\\\\\frac{V}{h} = 4\\\\\\\\pi\\\\\\\\left(\\\\\\\\frac{h}{2}\\\\\\\\right)^2\\\\\\\$, and taking in account that \\\\\\\$\\\\\\\\rho = \\\\\\\\frac{q_{enc}}{V}\\\\\\\$
+Using Gauss’s Law for the sphere at its surface ($r = h/2$):
+$$E_A \cdot 4\pi \left(\frac{h}{2}\right)^2 = \frac{Q_{enc}}{\varepsilon_0}$$
−#### Answer 1
+Given $V = \frac{4}{3}\pi\left(\frac{h}{2}\right)^3$ and $Q_{enc} = \rho \cdot V$:
+$$E_A \cdot 4\pi \left(\frac{h}{2}\right)^2 = \frac{\rho \cdot \frac{4}{3}\pi\left(\frac{h}{2}\right)^3}{\varepsilon_0}$$
−\\\\\\\$\\\\\\\$
−E_A = \\\\\\\\frac{\\\\\\\\rho h}{6\\\\\\\\varepsilon_0}
−\\\\\\\$\\\\\\\$
+**Answer 1:**
+$$E_A = \frac{\rho h}{6\varepsilon_0}$$
−For point \\\\\\\$B\\\\\\\$, the way is similar: in this case the plate without cavity generates certain field in \\\\\\\$B\\\\\\\$, located on its surface. From Gauss's Law
+---
−\\\\\\\$\\\\\\\$
−E_p\\\\\\\\cdot S = \\\\\\\\frac{q_t}{\\\\\\\\varepsilon_0}
−\\\\\\\$\\\\\\\$
+#### Field at Point B
+Point $B$ is on the surface of the plate.
+1. **Field from the solid plate ($E_p$):** By Gauss’s Law, the total field emerging from both sides of a plate is $\frac{\rho h}{\varepsilon_0}$. For just one side (at point $B$):
+ $$E_{plate} = \frac{\rho h}{2\varepsilon_0}$$
+2. **Field from the "negative" sphere ($E_s$):** Point $B$ is on the surface of the sphere ($r = h/2$). As calculated in Part 1, the field from the sphere at its surface is:
+ $$E_{sphere} = \frac{\rho h}{6\varepsilon_0}$$
−as \\\\\\\$S = \\\\\\\\frac{V}{h}\\\\\\\$, and \\\\\\\$\\\\\\\\rho = \\\\\\\\frac{q_t}{V}\\\\\\\$
+The vectors point in opposite directions at $B$, so we subtract them:
+$$E_B = E_{plate} - E_{sphere} = \frac{\rho h}{2\varepsilon_0} - \frac{\rho h}{6\varepsilon_0}$$
−\\\\\\\$\\\\\\\$
−E_p = \\\\\\\\frac{\\\\\\\\rho h}{\\\\\\\\varepsilon_0}
−\\\\\\\$\\\\\\\$
+**Answer 2:**
+$$E_B = \frac{\rho h}{3\varepsilon_0}$$
−but, for one of the side of plate, field is \\\\\\\$E_B' = \\\\\\\\frac{E_p}{2} = \\\\\\\\frac{\\\\\\\\rho h}{2\\\\\\\\varepsilon_0}\\\\\\\$ Now, overlapping with field generated by sphere,
+---
−\\\\\\\$\\\\\\\$
−E_B = E_B' - E_c
−\\\\\\\$\\\\\\\$
+#### Field along line OA
+Since line $OA$ is at the center of the infinite plate, the plate's own contribution to the field is zero at every point along this line. Thus, the field depends only on the charged sphere.
−where \\\\\\\$E_c = E_A\\\\\\\$,
+For any point at a distance $r$ inside the cavity ($0 \leq r \leq h/2$), we apply Gauss's Law to the sphere:
+$$E(r) \cdot 4\pi r^2 = \frac{\rho \cdot \frac{4}{3}\pi r^3}{\varepsilon_0}$$
−#### Answer 2
−
−\\\\\\\$\\\\\\\$
−E_B = \\\\\\\\frac{\\\\\\\\rho h}{3\\\\\\\\varepsilon_0}
−\\\\\\\$\\\\\\\$
−
−As \\\\\\\$A\\\\\\\$ is on line \\\\\\\$OA\\\\\\\$, all points over that line "feel" only the electric field generated by sphere. From Gauss's Law
−
−\\\\\\\$\\\\\\\$
−-E(r) \\\\\\\\cdot 4\\\\\\\\pi r^2 = -\\\\\\\\frac{q_{enc}}{\\\\\\\\varepsilon_0}
−\\\\\\\$\\\\\\\$
−
−as \\\\\\\$V = \\\\\\\\frac{4}{3}\\\\\\\\pi r^3\\\\\\\$, so \\\\\\\$3\\\\\\\\frac{V}{r} = 4\\\\\\\\pi r^2\\\\\\\$, and \\\\\\\$\\\\\\\\rho = \\\\\\\\frac{q_{enc}}{V}\\\\\\\$
−
−#### Answer 3
−
−\\\\\\\$\\\\\\\$
−E(r) = \\\\\\\\frac{\\\\\\\\rho r}{3\\\\\\\\varepsilon_0}
−\\\\\\\$\\\\\\\$
−
−for \\\\\\\$0\\\\\\\\leq r \\\\\\\\leq \\\\\\\\frac{h}{2}\\\\\\\$.
−
−
−
−
−
−
+**Answer 3:**
+$$E(r) = \frac{\rho r}{3\varepsilon_0}$$