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What is the strength of the electric field between two parallel conducting plates separated by 6 cm and having a potential difference (voltage) between them of 4.18 ×10^4 V ? Give answer in terms of 10^6 V/m.

Answer :

jorgezarate

Answer:

[tex]E=0.697*10^{6}V/m\\[/tex]

Explanation:

If the voltage is constant, the relation between the electric field E and the voltage V , for two plates separated by a distance d, is:

[tex]V=E*d\\[/tex]

d=6cm=0.06m

V=4.18×10^4 V

We solve to find E:

[tex]E=V/d=4.18*10^{4}/0.06=6.97*10^{5}=0.697*10^{6}V/m\\[/tex]

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