A15. μf capacitor is connected to a 50. v battery and becomes fully charged. the battery is removed and a slab of dielectric that completely fills the space between the plates is inserted. if the dielectric has a dielectric constant of 5.0, what is the voltage across the capacitor's plates after the slab is inserted?

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Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge,    then the following gives us the potential difference,      the formula  may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get    using q = 8.00  × 10⁻¹⁹ c and  δv = 2.00 v, we get    the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.

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