What is the ratio E2 E1 of the electric fields at the “surfaces” of the two spheres once equi- librium has been reached?
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Consider two "solid" conducting spheres with radii r1 = 7R and r2 = 8R , separated by a large distance so that the field and the potential at the surface of sphere #1 only depends on the charge on #1 and the corresponding quantities on #2 only depend on the charge on #2. Place an equal amount of charge on both spheres: q1 = q2 = Q. Now "connect" the two spheres with a wire. There will be a flow of charge through the wire until equilibrium is established. What is the ratio E2/E1 of the electric fields at the "surfaces" of the two spheres once equilibrium has been reached? What is the charge q1 on sphere #1?
Adi S.
An electrically neutral, conducting sphere with radius R is placed close to a large plane sheet of surface charge density +σ. Under the condition of electrostatic equilibrium, the electric field E at the center 0 produced by the induced charges on the sphere is:
Vishal G.
Given charge distribution on the surface $\sigma=\vec{a} \cdot \vec{r}$ is shown in the figure. Symmetry of this distribution implies that the sought $\vec{E}$ at the centre $O$ of the sphere is opposite to $\vec{a}$ $d q=\sigma(2 \pi r \sin \theta) r d \theta=(\vec{a} \cdot \vec{r}) 2 \pi r^{2} \sin \theta d \theta=2 \pi a r^{3} \sin \theta \cos \theta d \theta$ Again from symmetry, field strength due to any ring element $d E$ is also opposite to $\vec{a}$ i.e. $d \vec{E} \uparrow \downarrow \vec{a}$. Hence $d \vec{E}=\frac{d q r \cos \theta}{4 \pi \varepsilon_{0}\left(r^{2} \sin ^{2} \theta+r^{2} \cos ^{2} \theta\right)^{3 / 2}} \frac{-\vec{a}}{a}$ (Using the result of 3.9) $=\frac{\left(2 \pi a r^{3} \sin \theta \cos \theta d \theta\right) r \cos \theta}{4 \pi \varepsilon_{0} r^{3}} \frac{(-\vec{a})}{a}$ $=\frac{-\overrightarrow{a r}}{2 \varepsilon_{0}} \sin \theta \cos ^{2} d \theta$ Thus $\vec{E}=\int d \vec{E}=\frac{(-\vec{a}) r}{2 \varepsilon_{0}} \int_{0}^{\pi} \sin \theta \cos ^{2} \theta d \theta$ Integrating, we get $\vec{E}=-\frac{\overrightarrow{a r}}{2 \varepsilon_{0}} \frac{2}{3}=-\frac{\vec{a} r}{3 \varepsilon_{0}}$
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