### Correct Option: A. Increase ### Detailed Explanation: When a rectangular glass block is inserted between the plates of a parallel plate capacitor, the capacitance of the capacitor changes due to the properties of the dielectric material (in this case, glass) that is introduced into the capacitor's electric field. #### Step-by-Step Explanation: 1. **Understanding Capacitance**: - The capacitance \( C \) of a parallel plate capacitor is given by the formula: \[ C = \frac{\varepsilon A}{d} \] where: - \( C \) is the capacitance, - \( \varepsilon \) is the permittivity of the material between the plates, - \( A \) is the area of the plates, - \( d \) is the distance between the plates. 2. **Effect of Inserting a Dielectric**: - When a dielectric material (like glass) is inserted between the plates, the permittivity \( \varepsilon \) increases. The permittivity of a dielectric is given by: \[ \varepsilon = \varepsilon_0 \cdot \kappa \] where: - \( \varepsilon_0 \) is the permittivity of free space (vacuum), - \( \kappa \) (kappa) is the dielectric constant of the material (for glass, \( \kappa \) is greater than 1). 3. **Capacitance with Dielectric**: - The new capacitance \( C' \) when the dielectric is inserted becomes: \[ C' = \frac{\varepsilon' A}{d} = \frac{\kappa \varepsilon_0 A}{d} \] - Since \( \kappa > 1 \), it follows that \( C' > C \). Therefore, the capacitance increases when the dielectric is inserted. 4. **Physical Interpretation**: - The dielectric material reduces the electric field strength between the plates for a given charge, allowing the capacitor to store more charge at the same voltage. This is because the dielectric polarizes in the presence of the electric field, which effectively reduces the field strength and allows more charge to accumulate on the plates. ### Why Other Options Are Incorrect: - **Option B: Decrease**: This option is incorrect because inserting a dielectric increases the capacitance, not decreases it. A decrease would imply that the ability to store charge has diminished, which contradicts the fundamental behavior of dielectrics. - **Option C: Decrease then Increase**: This option suggests a transient behavior that does not occur in this scenario. The capacitance does not decrease initially and then increase; it simply increases upon the insertion of the dielectric. - **Option D: Remain Constant**: This option is incorrect because the introduction of a dielectric changes the permittivity of the space between the plates, which directly affects the capacitance. The capacitance cannot remain constant if the dielectric constant changes. ### Common Pitfalls: - **Misunderstanding Dielectric Effects**: Students often confuse the effects of dielectrics with other factors affecting capacitance, such as plate area or distance. Remember, inserting a dielectric always increases capacitance if the area and distance remain constant. - **Forgetting the Role of Dielectric Constant**: Not recognizing that the dielectric constant \( \kappa \) is always greater than 1 for materials like glass can lead to incorrect conclusions about capacitance changes. ### Revision Summary: - Inserting a dielectric (like glass) between capacitor plates increases capacitance. - Capacitance formula: \( C = \frac{\varepsilon A}{d} \) where \( \varepsilon \) increases with a dielectric. - Dielectric materials allow capacitors to store more charge at the same voltage. - Always remember that the dielectric constant \( \kappa \) is greater than 1, leading to increased capacitance.