6. A charged particle of 5.1 nC and another with 2.6 nC are separated by a distance of 5.5 m. What is the electric potential energy between the particles?

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kollybaba55 kollybaba55 · Answer 1 · 2020-05-18T02:17:00+02:00

Answer:

The electric potential energy between the two particles is [tex]U_{E} = 2.17 * 10^{-8} J[/tex]

Explanation:

[tex]q_{1} = 5.1 nC = 5.1 * 10^{-9} C\\q_{2} = 2.6 nC = 2.6 * 10^{-9} C\\r = 5.5 m[/tex]

Electric potential energy between two particles is given by the formula:

[tex]U_{E} = \frac{kq_{1}q_{2} }{r}\\U_{E} = \frac{9 * 10^9 * 5.1 * 10^{-9} * 2.6 * 10^{-9} }{5}\\U_{E} = 2.17 * 10^{-8} J[/tex]

facundo3141592 facundo3141592 · Answer 2 · 2020-05-18T02:20:40+02:00

Answer: Ue = 2.17x10^-8 N*m

Explanation:

The equation for the electric potential charge is:

Ue = k*q1*q2/r

where q1 and q2 are the charges, k is a constant k = 9x10^9 N.m^2/C^2

Here we have coulombs, and in the charges we have nanocoulombs, so we need to do a change of units.

q1 = 5.1 nC = 5.1x10^-9 C

q2 = 2.6 nC = 2.6x10^-9 C

and r is the distance between the charges, in this case r = 5.5m

Then we have:

Ue = (9x10^9 Nm^2/C^2)*(2.6x10^-9 C)*(5.1x10^-9 C)/5.5 m

Ue = (9*5.1*2.6/5.5)x10^-9 N*m

Ue = 21.7x10^-9 N*m

Ue = 2.17x10^-8 N*m