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A uniform electric field, with a magnitude of 370 N/C, is directed parallel to the positive x-axis. If the electric potential at x = 2.00 m is 1 000 V, what is the change in potential energy of a particle with a charge of + 2.80 x 10-3 C as it moves from x = 1.9 m to x = 2.1 m?

Respuesta :

Answer:

[tex]\Delta U = 0.2072 J[/tex]

Explanation:

Potential difference between two points in constant electric field is given by the formula

[tex]\Delta V = E.\Delta x[/tex]

here we know that

[tex]E = 370 N/C[/tex]

also we know that

[tex]\Delta x = 2.1 - 1.9 = 0.2 m[/tex]

now we have

[tex]\Delta V = 370 (0.2) = 74 V[/tex]

now change in potential energy is given as

[tex]\Delta U = Q\Delta V[/tex]

[tex]\Delta U = (2.80 \times 10^{-3})(74)[/tex]

[tex]\Delta U = 0.2072 J[/tex]

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