Answer:
Explanation:
Electric field due to a charge Q at a point d distance away is given by the expression
E = k Q / d , k is a constant equal to 9 x 10⁹
Field due to charge = 3 X 10⁻⁹ C
E = E = 
Field due to charge = 4 X 10⁻⁹ C
![E = [tex]\frac{9\times 10^9\times4\times10^{-9}}{(2-d)^2}](https://tex.z-dn.net/?f=E%20%3D%20%5Btex%5D%5Cfrac%7B9%5Ctimes%2010%5E9%5Ctimes4%5Ctimes10%5E%7B-9%7D%7D%7B%282-d%29%5E2%7D)
These two fields will be equal and opposite to make net field zero
= ![[tex]\frac{9\times 10^9\times4\times10^{-9}}{(2-d)^2}](https://tex.z-dn.net/?f=%5Btex%5D%5Cfrac%7B9%5Ctimes%2010%5E9%5Ctimes4%5Ctimes10%5E%7B-9%7D%7D%7B%282-d%29%5E2%7D)
[/tex]
d = 0.928
gravitational potential is directly proportional to the height of the object relative to a reference line and is given as
PE = mgh
where m = mass of object , g = acceleration due to
gravity and h = height of the object above the reference line .
as the skydiver falls , its height above the ground decrease and hence the gravitational potential energy of the skydiver decrease.
as per conservation of energy , total energy of the skydiver must remain constant all the time . hence the decrease in potential energy appears as increase in kinetic energy by same amount to keep the total energy constant
KE + PE = Total energy
so as the skydiver falls , it gains speed and hence the kinetic energy of skydiver increase since kinetic energy is directly proportional to the square of the speed.
when the parachute opens, the skydiver experience force in upward which tries to balance the weight of the skydiver. hence the speed of the skydiver decrease until upward force becomes equal to the downward force. hence the kinetic energy decrease just after the parachute opens
