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IrinaVladis [17]

3 years ago

11

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.

1 answer:

Vesna [10]3 years ago

3 0

Answer:

16.9000000000000001 J

Explanation:

From the given information:

Let the initial kinetic energy from point A be $K_A$ = 1.9000000000000001 J

and the final kinetic energy from point B be $K_B$ = ???

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

$-Q \times ( V_B - V_A) = (K_B - K_A)$

$-(6\times 10^{-3}\ C) \times ( -2.5 \times 10^3) = (K_B - 1.9000000000000001 \ J)$

$15 = (K_B - 1.9000000000000001 \ J)$

$K_B = 15+ 1.9000000000000001 \ J$

$\mathbf{K_B =1 6.9000000000000001 \ J}$

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Let L be the total length of wire and L' be the length of short piece of wire.

We know that

$R=\frac{\rho L}{A}$ = $\frac{\rho}{A}L=KL$

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The resistance of the long piece= $\frac{91}{8}\Omega$

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