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A parallel plate capacitor has a potential difference between its plates of 1.2 V and a plate separationdistance of 2.0 mm. What is the magnitude of the electric field if a material that has a dielectric constant of 3.3 is inserted between the plate

Respuesta :

Answer:

[tex]E_{med} = 181.8 V/m[/tex]

Explanation:

Potential difference between the plates of capacitor is given as

[tex]\Delta V = 1.2 V[/tex]

distance between the plates is given as

[tex]d =  2mm[/tex]

so now we have

[tex]E . d = \Delta V[/tex]

[tex]E = \frac{\Delta V}{d}[/tex]

[tex]E = \frac{1.2}{0.002}[/tex]

[tex]E = 600 V/m[/tex]

now the dielectric of constant 3.3 is inserted between the plates

so we know that

[tex]E_{med} = \frac{E}{k}[/tex]

so we have

[tex]E_{med} = \frac{600}{3.3}[/tex]

[tex]E_{med} = 181.8 V/m[/tex]

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