.012 J is going to be your answer
Answer:
Explanation:
We shall apply Gauss's theorem for electric flux to solve the problem . According to this theorem , total electric flux coming out of a charge q can be given by the following relation .
∫ E ds = q / ε
Here q is assumed to be enclosed in a closed surface , E is electric intensity on the surface so
∫ E ds represents total electric flux passing through the closed surface due to charge q enclosed in the surface .
This also represents total flux coming out of the charge q on all sides .
This is equal to q / ε where ε is a constant called permittivity which depends upon the medium enclosing the charge . For air , its value is 8.85 x 10⁻¹² .
If charge remains the same but radius of the sphere enclosing the charge is doubled , the flux coming out of charge will remain the same .
It is so because flux coming out of charge q is q / ε . It does not depend upon surface area enclosing the charge . It depends upon two factors
1 ) charge q and
2 ) the permittivity of medium ε around .
Answer: False
Explanation: In order to explain this problem we have to use the Faraday law, which say
dФm/dt=-ε it means that the variation of the magnetic field flux with time is equal to the emf ( electromotive force). In our case the magnetic flux is constant then there is not a emf induced in a wire closed loop.
Answer:
this makes no since so i cant help you here sorry
Answer:
The electric field is 
and the ditection is from outer to inner side of the membrane.
Explanation:
We know the electric field (
) is given by 
, 'V' being the potential.
In 1-D, it can be written as
where 'd' is the separation of space in between the potential difference is created.
Given, 
and the thickness of the cell membrane is 
.
Therefore the created electric field through the cell membrane is
