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The potential energy of a charge q moving from A to B at point A is 5.4 × 10-10 joules and at point B is 2.3 × 10-10 joules. The charge is 2.5 × 10-15 coulombs. What potential difference does the particle undergo?

Respuesta :

Balzano
Hi lamy
we know  ΔU=qΔV
solving for potential difference ΔV
we get ΔV=ΔU/q
ΔV= (Ub-Ua)/q
Ub is potential energy at b
Ua is potential energy at a
q is the charge of the particle
plug in and find out

Answer:

potential difference = 124000 Volts

Explanation:

As we know that potential energy per unit charge is known as potential of the point.

So here we can say that

[tex]V = \frac{U}{q}[/tex]

now we know that initial potential at point A is given as

[tex]V_a = \frac{U_a}{q}[/tex]

[tex]V_a = \frac{5.4 \times 10^{-10}}{2.5 \times 10^{-15}}[/tex]

[tex]V_a = 216000 Volts[/tex]

now similarly potential at point B is given as

[tex]V_b = \frac{U_b}{q}[/tex]

[tex]V_b = \frac{2.3 \times 10^{-10}}{2.5 \times 10^{-15}}[/tex]

[tex]V_b = 92000 Volts[/tex]

Now potential difference of two points is given as

[tex]\Delta V = V_a - V_b[/tex]

[tex]\Delta V = 216000 - 92000 = 124000 Volts[/tex]