
Expert Answer
Given that: A point positive charge $q$ is placed inside a spherical cavity with radius $R$ inside a conducting sphere. As per Gauss’s Law, the total electric flux is: \[\Phi_E=\oint\vec{E_f}\bullet\vec{ds}=\frac{q+q_{int}}{\varepsilon_o}\] Where: $q=$ Charge at the center $q_{int}=$ Charge induced on the inner surface Part (A) In a neutral sphere, the electrical field and flux are zero, hence: \[\Phi_E=\oint\vec{E_f}\bullet\vec{ds}=\frac{q+q_{int}}{\varepsilon_o}=0\] \[q+q_{int}=0\] \[q_{int}=-q\] Part (B) In a neutral sphere, the net charge across its internal and external surface will be zero. \[q_{int}{+q}_{ext}=0\] \[q_{ext}=-q_{int}\] \[q_{ext}=-(-q)=q\] Part (C) Let’s assume a point at a distance $r$ from the center of the cavity and $r<R$, so \[q_{enclosed}=q\] Hence: \[\Phi_E=\vec{E_f}\bullet A_{cavity}=\frac{q_{enclosed}}{\varepsilon_o}\] \[\vec{E_f}=\frac{q_{enclosed}}{{A_{cavity}\times\varepsilon}_o}\] \[\vec{E_f}=\frac{q}{{4\pi r^2\varepsilon}_o}=\frac{kq}{r^2}\]Numerical Result
The total charge $q_{int}$ that exists on the inside wall of the hollow sphere is “$-q$”. The total charge $q_{ext}$ that exists on the outside wall of the hollow sphere is “$q$”. The magnitude of the electric field $\vec{E_f}$ inside the hollow cavity is: \[\vec{E_f}=\frac{q}{{4\pi r^2\varepsilon}_o}=\frac{kq}{r^2}\]Example
A point charge is present at the center of the spherical cavity having a positive charge $q$. An external charge $q_2$ is brought near the conducting sphere. (A) Find the change in the total charge $q_{int}$ on the interior surface of the cavity. (B) Find the change in the total charge $q_{ext}$ on the exterior of the conductor. (C) Find the change in the Electric field $E_{cav}$ inside the cavity. (D) Find the change in the Electric field $E_{ext}$ outside the conductor. Solution Part (A) The total charge $q_{int}$ is only dependent on the positive charge $q$ placed in the center, hence it will not change in the presence of external charge $q_2$. Part (B) The total charge $q_{ext}$ is only dependent on the positive charge $q$ placed in the center, hence it will not change in the presence of external charge $q_2$. Part (C) Electric field $E_{cav}$ inside the cavity will not change because the conducting sphere is neutral and the electrical field inside the cavity is only dependent on the positive charge $q$. \[\vec{E_f}=\frac{q_{enclosed}}{{A_{cavity}\times\varepsilon}_o}\] Part (D) Electric field $E_{ext}$ outside the cavity will change because it will be distorted or affected by external charge $q_2$.Previous Question < > Next Question
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