Answer:
i) C decreases
ii) Q remains constant
iii) E remains constant
iv) ΔV increases
Explanation:
i)
We know, capacitance is given by:


<em>In this case as the distance between the plates increases the capacitance decreases while area and permittivity of free space remains constant.</em>
ii)
As the amount of charge has nothing to do with the plate separation in case of an open circuit hence the charge Q remains constant.
iii)
Electric field between the plates is given as:

where:
charge density, 
<em>As we know that distance of plate separation cannot affect area of the plate. Charge Q and permittivity are also not affected by it, so E remains constant.</em>
iv)
- From the basic definition of voltage we know that it is the work done per unit charge to move it through a distance.
- Here we increase the distance so the work done per unit charge increases.
The work done by the electric force to move a 1 c charge from a to b is
<em>(the voltage at 'b' compared to 'a') volts .</em>
<em></em>
That's because "volt" means "Joule per Coulomb" .
Positive will react better together. But opposites will try to get as far away as possible.
Shiddd ion even know but can you like my comment?
Answer:
E = 8.26*10⁻¹⁰ N/C, due south.
Explanation:
- Assuming no other forces acting on the electron than the electrostatic force due to the electric field, we can apply Newton's 2nd law as follows:
- Solving for E, we can find its magnitude as follows:

- The direction of the electric field is by definition the one that would take a positive test charge, so if the electron is accelerated to the north, the electric field would exactly oppose to this direction, so it is directed due south.