Question 1
\(C_o\) is the capacitance of a parallel plate capacitor with air as the dielectric. When half of the gap is filled with a dielectric of permittivity \(\epsilon_r\), then the modified capacitance is:
Solution
The original capacitance with air is given by:
Equation
\[C_o = \frac{A \epsilon_o}{d}\]
where:
- \(A =\) Area of the parallel plate capacitor
- \(d =\) Distance between the plates
When the gap is half-filled with a dielectric, the arrangement acts as two capacitors in parallel (since the potential difference is the same across both halves, but the area is split).
Let the two capacitors be \(C_1\) (air part) and \(C_2\) (dielectric part). Each occupies area \(A/2\).
Equation
\[C_1 = \frac{(A/2) \epsilon_o}{d} = \frac{A \epsilon_o}{2d} = \frac{C_o}{2}\]
Equation
\[C_2 = \frac{(A/2) \epsilon_o \epsilon_r}{d} = \frac{A \epsilon_o \epsilon_r}{2d} = \frac{C_o}{2} \epsilon_r\]
Since \(C_1\) and \(C_2\) are in parallel, the net capacitance is:
Equation
\[C_{net} = C_1 + C_2\]
Equation
\[C_{net} = \frac{A \epsilon_o}{2d} + \frac{A \epsilon_o \epsilon_r}{2d}\]
Equation
\[C_{net} = \frac{A \epsilon_o}{2d} (1 + \epsilon_r)\]
Substituting \(C_o = \frac{A \epsilon_o}{d}\):
Equation
\[C_{net} = \frac{C_o}{2} (1 + \epsilon_r)\]
Final Answer
Correct answer: A.