To determine the new capacitance when a thin metal plate is inserted halfway between the plates of a parallel plate capacitor, we first need to understand the effect of inserting a conductor between the plates of the capacitor.
Consider a parallel plate capacitor with capacitance \(C\). When a thin metal plate is inserted halfway between its plates without touching either side, essentially, two capacitors are formed in series, each with half the separation of the original capacitor.
Let's denote the original separation between the plates as \(d\) and the area of the plates as \(A\). The capacitance of a parallel plate capacitor is given by the formula:
\(C = \frac{{\varepsilon A}}{d}\)
After inserting the metal plate halfway (a distance \(d/2\) from each of the original capacitor plates), two new capacitors are formed in series:
The capacitance of each new capacitor can be calculated as:
\(C' = \frac{{\varepsilon A}}{d/2} = \frac{{2\varepsilon A}}{d} = 2C\)
Since these newfound capacitors are in series, their equivalent capacitance \(C_{\text{eq}}\) is given by:
\(\frac{1}{C_{\text{eq}}} = \frac{1}{C'} + \frac{1}{C'} = \frac{1}{2C} + \frac{1}{2C} = \frac{1}{C}\)
Thus, the equivalent capacitance \(C_{\text{eq}}\) is:
\(C_{\text{eq}} = C\)
Therefore, the correct answer is that the capacitance of the system remains \(C\) when the metal plate is inserted. The presence of the conducting plate in the middle, creating two capacitors in series each with double the capacitance, results in the overall capacitance remaining unchanged.
An important point to note here is that the metal plate does not alter the electric field but rather divides the system into two symmetric capacitors sharing the same charge, hence the original total capacitance is retained.
Therefore, the correct answer is C.
A 10 $\mu\text{C}$ charge is placed in an electric field of $ 5 \times 10^3 \text{N/C} $. What is the force experienced by the charge?