The plates of a parallel plate capacitor are \( d=2.0\,\mathrm{mm} \) apart and each plate has an area \( A=2.0\times10^{-4}\,\mathrm{m^{2}} \). The space between the plates is completely filled with a linear dielectric material of dielectric constant K = 2.0. The capacitor is connected to a source of emf \( 2.0\times10^{2}\,\mathrm{V} \). The electric field between the plates is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 1.0\times10^{5}\,\mathrm{V\,m^{-1}} \)
The capacitor stays connected to the source, so the potential difference across the plates is the emf V. The field between the plates is uniform, so \( V=Ed \). The dielectric raises the capacitance and the charge on the plates. But the field is still fixed by V and d, so the plate area is not needed.
In SI units the data are:
\( V=2.0\times10^{2}\,\mathrm{V} \)
\( d=2.0\,\mathrm{mm}=2.0\times10^{-3}\,\mathrm{m} \)
The electric field between the plates is:
\( \begin{aligned} E &= \frac{V}{d} \\ &= \frac{2.0\times10^{2}\,\mathrm{V}}{2.0\times10^{-3}\,\mathrm{m}} \\ &= 1.0\times10^{5}\,\mathrm{V\,m^{-1}} \end{aligned} \)