A pair of parallel conducting plates have potentials of −17 V and 55 V, respectively. An electron is ejected from the plate at potential −17 V and travels to the plate at 55 V. What is the speed of the electron as it strikes the 55 V plate, assuming it left the −17 V plate starting from rest? The mass of the electron is 9.11 × 10⁻³¹ kg.

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Quizplus · Accepted Answer

The speed of the electron can be found using the principle of conservation of energy. The potential energy difference between the two plates is converted into kinetic energy as the electron moves from the lower to the higher potential. The potential difference (V) is 55 V - (-17 V) = 72 V. The charge of an electron (e) is -1.6 × 10^-19 C (the negative sign indicates the charge of the electron but does not affect the calculation of energy since it will be squared in the kinetic energy formula or taken as absolute value in potential energy calculations). The initial kinetic energy is 0 since the electron starts from rest. The final kinetic energy (K.E.) is given by K.E. = 1/2 m v^2, where m is the mass of the electron (9.11 × 10^-31 kg) and v is the velocity of the electron. The potential energy (P.E.) difference is given by P.E. = eV, where V is the potential difference.Thus, eV = 1/2 m v^2.Rearranging for v gives v = sqrt((2eV)/m).Substituting the given values: v = sqrt((2 * 1.6 × 10^-19 C * 72 V) / (9.11 × 10^-31 kg)).Calculating this gives a velocity v ≈ 5.0 × 10^6 m/s.

A Pair of Parallel Conducting Plates Have Potentials of −17