Explanation:
The given data is as follows.

Voltage = 2.50 V
Hence, calculate the equivalence capacitor as follows.


= 
C = 
Now, we will calculate the charge across each capacitance as follows.
Q = CV
= 
=
=
Thus, we can conclude that
is the charge stored on each given capacitor.
Answer:
The distance from the charge is 3.35 m.
Explanation:
Given that,
Electric potential, V = 635 V
Magnitude of electric field, E = 189 N/C
We need to find the distance from the charge. We know that the relation between electric field and electric potential is given by :

d is the distance from charge

So, the distance from the charge is 3.35 m. Hence, this is the required solution.
Angle is 31 adding all sides
The velocity of the object. because the greater the movement in particles the greater would be the kinetic energy.