First slide

Motion of charged particle in uniform electric field

Question

An electron $(q = - 1.6 \times 10^{- 19}, m_{e} = 9.1 \times 10^{- 31} k g)$ is projected out along the + x axis with an initial speed of $2.0 \times 10^{6} m / s$ . It goes 20 cm and stops due to a uniform electric field in the region. Find the magnitude and direction of the field.

Moderate

A

50 N/C, +X direction
B

57 N/C, +X direction
C

60 N/C, -X direction
D

67 N/C, -X direction

Solution

$\begin{array}{l} V^{2} - u^{2} = 2 a s \\ - {(2.0 \times 10^{6})}^{2} = 2 a \times 20 \times 10^{- 2} \\ a = - 10^{13} m s^{- 2} \\ F = m a = 9.1 \times 10^{- 31} \times (- 10^{13}) = 9.1 \times 10^{- 18} N \\ E = \frac{F}{q} = \frac{9.1 \times 10^{- 18}}{1.6 \times 10^{- 19}} = 57 N / C (a p p r o x) \\ direction should be in +X direction \end{array}$

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