Question

1. A pair of charged conducting plates produces a uniform field of Eo = 10,859 N/C directed to the right, between the plates. The separation of the plates is L = 37 mm. In Figure, an electron (e = - 1.6 x 10-19 C; m = 9.1 x 10-31 kg) is projected from plate A, directly toward plate B, with an initial velocity of Vo = 2.1×107 m/s. The velocity of the electron (expressed in general form as a whole number) as it strikes plate B is what?

1.628*10^7 m/sec
1.523*10^6 m/sec
1.564*10^7 m/sec
1.731*10^7 m/sec

Answer 1

Answer on Question 74925, Physics, Other

Question:

1. A pair of charged conducting plates produces a uniform field of $E_0 = 10859 \, N / C$ directed to the right, between the plates. The separation of the plates is $L = 37 \, mm$ . In Figure, an electron ( $e = -1.6 \cdot 10^{-19} \, C$ ; $m_e = 9.1 \cdot 10^{-31} \, kg$ ) is projected from the plate $A$ , directly toward the plate $B$ , with an initial velocity of $v_0 = 2.1 \cdot 10^7 \, m/s$ . The velocity of the electron (expressed in general form as a whole number) as it strikes plate $B$ is what?

a) $1.628 \cdot 10^{7} \, m/s$

b) $1.523 \cdot 10^{6} \, m/s$

c) $1.564 \cdot 10^{7} \, m/s$

d) $1.731 \cdot 10^{7} \, m/s$

Solution:

We can find the velocity of the electron as it strikes the plate $B$ from the work-kinetic energy theorem (the work done by the electric field against the electron is equal to the change in the kinetic energy):

$$W = K E _ {f i n a l} - K E _ {i n i t i a l},$$ $$F _ {e} L = \frac {1}{2} m _ {e} v ^ {2} - \frac {1}{2} m _ {e} v _ {0} ^ {2},$$

here, $F_{e} = eE_{0}$ is the electric force acting on the electron, $v_{0}$ is the initial velocity of the electron, $v$ is the velocity of the electron as it strikes plate $B$ , $e$ is the charge of the electron, $m_{e}$ is the mass of the electron, $E_{0}$ is the magnitude of the uniform electric field, $L$ is the separation of the plates.

Then, we can write:

$$e E _ {0} L = \frac {1}{2} m _ {e} v ^ {2} - \frac {1}{2} m _ {e} v _ {0} ^ {2},$$ $$2eE_0L = m_e v^2 - m_e v_0^2,$$ $$v^2 = v_0^2 + \frac{2eE_0L}{m_e},$$ $$\begin{aligned} v &= \sqrt{v_0^2 + \frac{2eE_0L}{m_e}} \\ &= \sqrt{\left(2.1 \cdot 10^7 \frac{m}{s}\right)^2 + \frac{2 \cdot (-1.6 \cdot 10^{-19} C) \cdot 10859 \frac{N}{C} \cdot 0.037 \, m}{9.1 \cdot 10^{-31} \, kg}} \\ &= 1.731 \cdot 10^7 \, \frac{m}{s}. \end{aligned}$$

Answer:

d) $v = 1.731 \cdot 10^7 \, \text{m/s}$.

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