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Correct Answer: c = (ε0a)/ d
to properly explain how the capacitance of a parallel plate capacitor changes when a dielectric material is inserted, let's first revisit the basic formula for the capacitance of a parallel plate capacitor without a dielectric. the capacitance, \( c \), is given by:
\[ c = \frac{\epsilon_0 a}{d} \]
where:
- \( \epsilon_0 \) is the permittivity of free space,
- \( a \) is the area of one of the plates, and
- \( d \) is the separation between the two plates.
now, when a dielectric material is inserted between the plates of the capacitor, the capacitance increases. this is because the dielectric reduces the electric field within the capacitor for a given charge, thereby increasing the capacitor's ability to store charge. the capacitance of a capacitor with a dielectric can be calculated using the modified formula:
\[ c = \frac{\epsilon a}{d} \]
here, \( \epsilon \) represents the permittivity of the dielectric material, which is the product of the permittivity of free space \( \epsilon_0 \) and the relative permittivity (or dielectric constant) \( \kappa \) of the material:
\[ \epsilon = \kappa \epsilon_0 \]
therefore, the formula becomes:
\[ c = \frac{\kappa \epsilon_0 a}{d} \]
this shows that the capacitance \( c \) is increased by a factor of \( \kappa \), the dielectric constant of the material inserted. the dielectric constant is always greater than 1, hence the capacitance with a dielectric material is always greater than the capacitance without it.
in summary, inserting a dielectric material between the plates of a parallel plate capacitor increases its capacitance by reducing the effective electric field and thereby allowing the capacitor to store more charge for the same voltage. the correct formula to use when a dielectric is present is:
\[ c = \frac{\kappa \epsilon_0 a}{d} \]
this formula accounts for the effect of the dielectric in increasing the capacitance of the capacitor.
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