Question

# In the uniform electric field of $$E = 1 \times {10}^{4} N{C}^{-1}$$, an electron is accelerated from rest. The velocity of the electron when it has travelled a distance of $$2 \times {10}^{-2} m$$ is nearly ________ $$m{s}^{-1}$$  $$(\dfrac{e}{m}$$ of electron $$\simeq 1.8 \times {10}^{11} C {kg}^{-1}$$)

A
1.6×106
B
0.85×106
C
0.425×106
D
8.5×106

Solution

## The correct option is D $$8.5 \times {10}^{6}$$Given : Distance covered   $$S = 2\times 10^{-2}$$ mAcceleration of the electron    $$a = \dfrac{eE}{m}$$$$\therefore$$  $$a = 1.8\times 10^{11}\times 10^4 = 1.8\times 10^{15}$$  $$m/s^2$$Initial speed of the electron is zero  i.e. $$u = 0$$Using    $$v^2 -u^2 = 2aS$$$$\therefore$$    $$v^2 - 0 = 2(1.8\times 10^{15})(2\times 10^{-2})$$          $$\implies v = 8.5\times 10^{6}$$  $$m/s$$Physics

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