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Rzqust [24]
4 years ago
4

Two flat 4.0 cm × 4.0 cm electrodes carrying equal but opposite charges are spaced 2.0 mm apart with their midpoints opposite ea

ch other. between the electrodes but not near their edges, the electric field strength is 6.8 × 106 n/c. what is the magnitude of the charge on each electrode? (ε 0 = 8.85 × 10-12 c2/n · m2)
Physics
1 answer:
Anastaziya [24]4 years ago
3 0

Answer:

9.5 \cdot 10^{-8} C

Explanation:

The electric field between two charged plates is uniform, and its strength is given by

E=\frac{\sigma}{\epsilon_0}

where

\sigma=\frac{Q}{A} is the charge density, with Q being the charge on one plate and A the area of the plate

\epsilon_0=8.85 \cdot 10^{-12} C^ N^{-1} m^{-2} is the dielectric constant

Since we know the magnitude of the field, E=6.8 \cdot 10^6 N/C, we can find the charge density:

\sigma=\epsilon_0 E=(8.85 \cdot 10^{-12})(6.8 \cdot 10^6 )=6.0 \cdot 10^{-5} C m^{-2}

The area of each plate is

A=(0.04 m)^2=1.6 \cdot 10^{-3} m^2

So the charge on each electrode is

Q=\sigma A=(6.0 \cdot 10^{-5})(1.6 \cdot 10^{-3})=9.5 \cdot 10^{-8} C

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Answer:

i)20369 photons

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Explanation:

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7 0
3 years ago
Need help asap pls
ozzi
Detailed Explanation:

1) Rusting of Iron

4Fe + 3O2 + 2H2O -> 2Fe2O32H2O

Reactants :-
Fe = 4
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Products :-
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O = 2 * 3 + 2 = 8
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2) Fermentation of sucrose…

C12H22O11 + H2O -> 4C2H5OH + 4CO2

Reactants :-
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Products :-
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H = 4 * 5 + 4 = 24
O = 4 * 2 + 4 = 12

Looking closely at the way I have taken the total number of elements on the reactants and products side, you can solve the rest.

All the Best!
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3 years ago
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Answer:

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The difference between these two values ​​is less than 1.2%

for smaller angle the difference is reduced more

Therefore, the period for both the 5º and 10º angles is almost the same

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