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A particle with charge 6 mC moving in a region where only electric forces act on it has a kinetic energy of 1.9000000000000001 J at point A. The particle subsequently passes through point B which has an electric potential of -2.5 kV relative to point A. Determine the kinetic energy of the particle as it moves through point B.

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ajeigbeibraheem

Answer:

16.9000000000000001 J

Explanation:

From the given information:

Let the initial kinetic energy from point A be [tex]K_A[/tex] = 1.9000000000000001 J

and the final kinetic energy from point B be [tex]K_B[/tex] = ???

The charge particle Q = 6 mC = 6 × 10⁻³ C

The change in the electric potential from point B to A;

i.e. V_B - V_A = -2.5 × 10³ V

According to the work-energy theorem:

-Q × ΔV = ΔK

[tex]-Q \times ( V_B - V_A) = (K_B - K_A)[/tex]

[tex]-(6\times 10^{-3}\ C) \times ( -2.5 \times 10^3) = (K_B - 1.9000000000000001 \ J)[/tex]

[tex]15 = (K_B - 1.9000000000000001 \ J)[/tex]

[tex]K_B = 15+ 1.9000000000000001 \ J[/tex]

[tex]\mathbf{K_B =1 6.9000000000000001 \ J}[/tex]

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