Answer:
e.)At twice the distance, the strength of the field is E/4.
Step-by-step explanation:
The strength of the electric field at a certain distance from a point charge is given by:
![E=k(Q)/(r^2)](https://img.qammunity.org/2020/formulas/physics/high-school/dlapwc2owjbl2muv9l2hawq453hgc50k7g.png)
where
k is the Coulomb's constant
Q is the charge
r is the distance from the point charge
In this problem, the distance from the point charge is doubled:
r' = 2r
So the new electric field strength is
![E'=k(Q)/((2r)^2)=k (Q)/(4 r^2)=(1)/(4) (k(Q)/(r^2))=(E)/(4)](https://img.qammunity.org/2020/formulas/physics/high-school/m010yf9t78kwp55kllckrfrmsmjtuxd94p.png)
so, at twice the distance the strength of the field is E/4.