Follow two steps below to derive Eq. (6): i. Insert the angular acceleration from Eq. (2) and the moment of inertia from Eq. (5) into RHS of Eq. (4) and find the friction in terms of these quantities. ii. Insert the friction obtained in i. into Eq. (3) and rearrange it to get Eq. (6)

spare some change and give it to the poor

explanation:

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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