Question

An electron is released from rest in a weak electric field given by E,--2.50 x 10.10 N/C j. After the electron has traveled a vertical distance of 1.8 μm, what is its speed?

Answer 1

Answer:

$v = 1.26 \times 10^8 m/s$

Explanation:

Force on an electron while it moves through constant electric field is given as

$F = eE$

$e = 1.6 \times 10^{-19} C$

$E = 2.50 \times 10^{10} N/C$

$F = (2.50 \times 10^{10})(1.6 \times 10^{-19})$

$F = 4 \times 10^{-9} N$

now by work energy theorem we can say

Work done by electric field = kinetic energy of the electron

$F . d = \frac{1}{2}mv^2$

$(4 \times 10^{-9})(1.8 \times 10^{-6}) = \frac{1}{2}(9.11 \times 10^{-31})v^2$

$v = 1.26 \times 10^8 m/s$