Question

Two metallic terminals protrude from a device. The terminal on the left is the positive reference for a voltage called vx (the other terminal is the negative reference). The terminal on the right is the positive reference for a voltage called vy (the other terminal being the negative reference). If it takes 1 mJ of energy to push a single electron into the left terminal, determine the voltages vx and vy.

Answer 1

Answer:

vx=6.242PV

vy=-6.242PV

Explanation:

Two metallic terminals protrude from a device. The terminal on the left is the positive reference for a voltage called vx (the other terminal is the negative reference). The terminal on the right is the positive reference for a voltage called vy (the other terminal being the negative reference). If it takes 1 mJ of energy to push a single electron into the left terminal, determine the voltages vx and vy.

Here 1mj of energy is needed to an electron into the other terminal

voltage of an electron

Recall that E=IVt(product of power and time

but Q=quantity of charge is it     (measured in coulombs)

E/Q=V

Ve=$\frac{-1*10^{-3} }{-1.602*10^{-19} }$

0.6242*10^16

6.242*10^15

6.242PV(P=Peta=10^15)

voltage ve=6.242PV

Ve is used as Vy

The voltage vf is

Vf=$\frac{1*10^{-3} }{-1.602*10^{-19} }$

Vf=-6.242*10^15

vf=-6.242PV

therefore, vf is designated as vy

the voltages are of the same magnitude but in different direction

Answer 2

Answer:

Vx = 6.242 x 10raised to power 15

Vy = -6.242 x 10raised to power 15

Explanation:

from E = IVt

but V = IR from ohm's law and Q = It from faraday's first law

I = Q/t

E = Q/t x V x t = QV

hence, E =QV

V = E/Q