Parallel Plate Capacitor: Practice Problem & Solution
A capacitor of capacitance \(C\) is charged to a potential \(V\). The flux of the electric field through a closed surface enclosing the capacitor is
Solution Explained:
To solve this problem, we apply the core principles of Parallel Plate Capacitor. Understanding the underlying formula is key to arriving at the correct answer below:
A capacitor consists of two plates with equal and opposite charges, \(+Q\) and \(-Q\), where \(Q = CV\). If a closed surface encloses the entire capacitor, the net charge enclosed within the surface is \(Q_{\text{enclosed}} = (+Q) + (-Q) = 0\). By Gauss's Law, the total electric flux through this closed surface is \(Phi_E = \frac{Q_{\text{enclosed}}}{\epsilon_0}\). Therefore, \(Phi_E = 0\).
Leave a Reply