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Two large parallel conducting plates separated by 6 cm carry equal and opposite surface charge densities such that the electric field between them is uniform. The difference in potential between the plates is 400 V. An electron is released from rest at the negatively charged plate.
(a) What is the magnitude of the electric field between the plates?
(b) Which plate is at the higher potential?
1. The negatively charged plate
2. the positively charged plat

Respuesta :

Answer:

a. 6666.67 V/m

b. 2. the positively charged plat

Explanation:

a. The computation of the magnitude of the electric field between the plates is shown below:

As we know that

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

[tex]= \frac{400 V}{0.06 m}[/tex]

= 6666.67 V/m

hence, the magnitude of the electric field is 6666.67 V/m

b. Based on this the higher potential is positively charged plate as the flow of the current goes from positive to negative and it is inverse in case of th electron

We simply applied the above formula  

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