Answer:
Ans 1 : The change of electric potential energy [tex]\Delta U= q(V_{2} -V_{1} )[/tex].
Ans 2 : Numerical value of the [tex]\Delta U = 131 eV[/tex]
Explanation:
Given :
The potential of one end [tex](V_{1} )[/tex] = 44 V
The potential of other end [tex](V_{2} )[/tex] = 175 V
We know that electric potential is the electric potential energy per unit charge.
Ans 1 :
So we write that, [tex]\Delta V = \frac{\Delta U}{q}[/tex]
∴ [tex]\Delta U = q \Delta U[/tex]
Here, electron moves from higher potential to lower potential.
∴ [tex]\Delta U = q (V_{2} -V_{1} )[/tex]
Ans 2 :
For numerical value we have to only subtract two potential,
⇒ [tex]\Delta U = (175 -44) eV[/tex]
[tex]\Delta U = 131eV[/tex]
Where [tex]e = 1.6 \times 10^{-19}[/tex] charge of electron.