
Since the flower pot is dropped, it has an initial velocity of zero. Also the flower pot accelerates due to gravity.
so
Conduction is the transfer of energy by collusion between the atoms and molecules in a material.
Answer:
The new force becomes 4 times the initial force.
Explanation:
The force of attraction or repulsion is given by the relation as follows :

Where
d is the distance between the interacting charges
F is inversely proportional to the distance between charges.
If the distance is halved, d'=(d/2), new force is given by :

So, the new force becomes 4 times the initial force.
Answer:
E = 12640.78 N/C
Explanation:
In order to calculate the electric field you can use the Gaussian theorem.
Thus, you have:

ФE: electric flux trough the Gaussian surface
Q: net charge inside the Gaussian surface
εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2
If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:

r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m
Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:

Finally, you obtain for E:

hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C