Answer:
the exposed core of a dead star, supported by electron degeneracy pressure.
Explanation:
A white dwarf is a low luminosity exposed core of a dead star having mass comparable to the sun but volume comparable to the earth . So its density is very high . These stars have lost the capacity to generate energy through the process of fusion . Due to high gravitational energy , it goes on shrinking but ultimately balanced by electron degeneracy pressure. It is not a main sequence star as it has lost the power of fusion .
Answer:
a)ΔV = 6.48 KV
b)ΔU =18.79 mJ
Explanation:
Given that
E= 1.8 KV/m
a)
We know that
Electric potential difference ΔV given as
ΔV = E .d
Here
E= 1.8 KV/m
d= 3.6 m
ΔV = E .d
ΔV = 1.8 x 3.6 KV
ΔV = 6.48 KV
b)
Given that
q=+2.90 µC
Change in electric potential energy ΔU given as
ΔU = q .ΔV

ΔU =18.79 mJ