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Charra [1.4K]
4 years ago
15

Two large parallel metal plates are 1.2 cm apart and have charges of equal magnitude but opposite signs on their facing surfaces

. Take the potential of the negative plate to be zero. If the potential halfway between the plates is then 6.6 V, what is the electric field in the region between the plates?
Physics
1 answer:
zheka24 [161]4 years ago
3 0

Answer:

E=1100V/m

                                                                     

Explanation:

Given           required           <u>solution</u>

V=6.6v         E=?                   V=Ed ;    V is the potential difference between                                                              

d=D/2=1.2cm/2=0.6cm=0.006m             the   halfway                  

                                                               E is the electric field between the two                                                                              

                                                                       plates.

                                                              d is the distance between the halfway.

So we can use the above formula to calculate the electric field.

    V=Ed   from this E=V/d  substitute the values from the given equation.

E=6.6v/0.006m

E= 1100 v/m

                                                                     

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Putting values in above equation, we get:

E=\frac{6.625\times 10^{-34}J.s\times 3\times 10^8m/s}{0.122m}\\\\E=1.63\times 10^{-24}J

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E=1.63\times 10^{-24}\times \frac{88.3}{100}=1.44\times 10^{-24}J

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Putting values in above equation, we get:

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n=\frac{q}{E}

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