An electron moving parallel to the x axis has an initial speed of 5.42 x 106 m/s at the origin. Its speed is reduced to 1.98 x 105 m/s at the point P at x = 2.00 cm. Calculate the electric potential change (in V) from the origin to point P.
Question
An electron moving parallel to the x axis has an initial speed of 5.42 x 106 m/s at the origin. Its speed is reduced to 1.98 x 105 m/s at the point P at x = 2.00 cm. Calculate the electric potential change (in V) from the origin to point P.
Solution
To solve this problem, we need to use the principle of conservation of energy. The initial kinetic energy of the electron is converted into electric potential energy as it moves from the origin to point P.
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First, calculate the initial kinetic energy (KE_initial) of the electron using the formula KE = 1/2 mv^2, where m is the mass of the electron (9.11 x 10^-31 kg) and v is the initial speed (5.42 x 10^6 m/s):
KE_initial = 1/2 * (9.11 x 10^-31 kg) * (5.42 x 10^6 m/s)^2
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Then, calculate the final kinetic energy (KE_final) of the electron at point P using the same formula, but with the final speed (1.98 x 10^5 m/s):
KE_final = 1/2 * (9.11 x 10^-31 kg) * (1.98 x 10^5 m/s)^2
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The change in kinetic energy is equal to the change in electric potential energy, which is given by the formula ΔPE = qΔV, where q is the charge of the electron (-1.6 x 10^-19 C) and ΔV is the change in electric potential. We can rearrange this formula to solve for ΔV:
ΔV = ΔPE / q
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Substitute the change in kinetic energy (KE_initial - KE_final) for ΔPE and the charge of the electron for q:
ΔV = (KE_initial - KE_final) / (-1.6 x 10^-19 C)
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Calculate the result to find the change in electric potential from the origin to point P. The negative sign indicates that the electric potential decreases as the electron moves from the origin to point P.
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