The electric field inside a cell membrane is 8.0 MN/C. Part A What's the force on a singly charged ion in this field?

Physics

1 answer:

dlinn [17]2 years ago

7 0

Answer:

Force on the singly charged ion will be $12.6\times 10^{-13}N$

Explanation:

We have given electric field E = 8 MN/C

So electric field in N/C will be $E=8MN/C=8\times 10^6N/C$

It is given that ion so charge on ion will be equal to $e=1.6\times 10^{-19}C$

We have to find the electric force on the ion

Electric force is equal to $F=qE$ , here q is charge and E is electric field

So force on the charge will be equal to $F=8\times 10^6\times 1.6\times 10^{-19}=12.6\times 10^{-13}N$

So force on the singly charged ion will be $12.6\times 10^{-13}N$

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An diagram of the seed's motion is attached to this solution.

This is very close to a projectile motion question. And the quantity to be calculated, how far along the grant a seed released would travel is called the Range.

And this would be obtained from the equations of motion,

First of, the height of the plant is related to some quantities of the motion with this relation.

H = u(y) t + 0.5g(t^2)

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(substituting the t = √(2H/g) derived from above

R = u(x) √(2H/g)

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QED!