A parallel-plate capacitor with air between its plates has capacitance C. If the space between the plates is completely filled with a dielectric material of dielectric constant 7, what will be the new capacitance?

Introduction / Context:
Capacitors store electric charge and energy, and their capacitance depends on geometry and the material between the plates. Replacing air with a dielectric material increases capacitance by a factor equal to the dielectric constant of that material. This concept is crucial in designing capacitors for electronic circuits, tuning circuits, and energy storage applications.

Given Data / Assumptions:
• Original capacitor: parallel plates with air as dielectric, capacitance C. • New capacitor: the same plates and separation, but the space is filled with a dielectric. • Dielectric constant (relative permittivity) of the material k = 7. • Plate area and separation remain unchanged.

Concept / Approach:
For a parallel-plate capacitor, capacitance with a dielectric fully filling the space between plates is given by C = k * epsilon0 * A / d, where k is the dielectric constant, epsilon0 is permittivity of free space, A is plate area, and d is plate separation. When air (k ≈ 1) is replaced by a dielectric with constant k, the capacitance increases by the factor k, assuming air is approximated as vacuum. Thus, if the original capacitance was C with air, the new capacitance with dielectric becomes k * C.

Step-by-Step Solution:
Step 1: Write the capacitance with air (or vacuum): C_air = epsilon0 * A / d. Step 2: The given capacitance C corresponds to C_air. Step 3: When a dielectric with constant k = 7 fully fills the space, capacitance becomes C_dielectric = k * epsilon0 * A / d. Step 4: Substitute C_air into this expression: C_dielectric = k * C_air = 7 * C. Step 5: Therefore, the new capacitance is 7C.

Verification / Alternative check:
If C = 2 microfarad with air, filling the space with a dielectric of k = 7 would give C_dielectric = 7 * 2 microfarad = 14 microfarad. This simple scaling behaviour is consistent with theory and with experimental use of dielectric materials to build high capacitance devices in smaller volumes.

Why Other Options Are Wrong:
Option a (C): This would mean the dielectric has no effect, which contradicts the basic definition of dielectric constant. Option b (C/7): This implies capacitance decreases, which is opposite to what happens when a dielectric is inserted. Option d (14C): This would require an effective dielectric constant of 14, not 7. Option e (C/14): This is also a decrease and has no basis in the formula for capacitance with a dielectric.

Common Pitfalls:
Students sometimes confuse relative permittivity (dielectric constant) with absolute permittivity epsilon0 or misremember whether the capacitance increases or decreases. A helpful reminder is that inserting a dielectric material increases the ability of the capacitor to store charge at the same voltage, so capacitance increases by the factor k. Be careful not to confuse this with effects on electric field strength or energy density, which involve related but different formulas.

Final Answer:
The capacitance of the capacitor becomes 7C when the dielectric with k = 7 completely fills the space between the plates.