A "gauge 8" jumper cable has a diameter d of 0.326 centimeters. The cable carries a current I of 30.0 amperes. The electric field E in the cable is 0.062 newtons per coulomb. part a: What electric field E′ would have been required to create a current of 30.0 amperes in a copper "gauge 10" wire with diameter d equal to 0.259 centimeters?

0.0979 N/c

Explanation:

Electric field, E is given as a product of resistivity and current density

E=jP where P is resistivity and j is current density

But the current density is given as

where I is current and A is area and

Substituting this into the first equation then

Given diameter of 0.259 cm= 0.00259 m and the radius will be half of it which is 0.001295 m

k2s for potassium sulphide

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j