Question

A capacitor of \(50\,\mu\mathrm{F}\) is connected to a \(200\ \mathrm{V}\) battery. A dielectric of \(K=4\) is inserted. Find the new capacitance and stored energy.

• A. \(200\,\mu\mathrm{F}\) and \(4\ \mathrm{J}\)
• B. \(100\,\mu\mathrm{F}\) and \(40\ \mathrm{J}\)
• C. \(4\,\mu\mathrm{F}\) and \(200\ \mathrm{J}\)
• D. \(40\,\mu\mathrm{F}\) and \(100\ \mathrm{J}\)

Hint:

If the battery remains connected, \[ \boxed{C'=KC} \] and \[ \boxed{U=\frac12 CV^2.} \]

Answer 1

The Correct Option is A

Step 1: Find the new capacitance. Since the battery remains connected, \[ C'=KC =4\times50\,\mu\mathrm{F} =200\,\mu\mathrm{F}. \]

Step 2: Calculate the stored energy. The energy stored is \[ U=\frac12 C'V^2. \] Substituting, \[ U=\frac12(200\times10^{-6})(200)^2 =4\ \mathrm{J}. \] Hence, \[ \boxed{C'=200\,\mu\mathrm{F},\qquad U=4\ \mathrm{J}.} \] Therefore, \[ \boxed{(A)} \] is the correct answer.