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zvonat [6]
3 years ago
14

A particle leaves the origin with a speed of 3 106 m/s at 38 degrees to the positive x axis. It moves in a uniform electric fiel

d directed along positive y axis. Find Ey such that the particle will cross the x axis at x
Physics
1 answer:
Salsk061 [2.6K]3 years ago
8 0

Answer:

If the particle is an electron E_y = 3.311 * 10^3 N/C

If the particle is a proton, E_y = 6.08 * 10^6 N/C

Explanation:

Initial speed at the origin, u = 3 * 10^6 m/s

\theta = 38^0 to +ve x-axis

The particle crosses the x-axis at , x = 1.5 cm = 0.015 m

The particle can either be an electron or a proton:

Mass of an electron, m_e = 9.1 * 10^{-31} kg

Mass of a proton, m_p = 1.67 * 10^{-27} kg

The electric field intensity along the positive y axis E_y, can be given by the formula:

E_y = \frac{2 m u^2 sin \theta cos \theta}{qx} \\

If the particle is an electron:

E_y = \frac{2 m_e u^2 sin \theta cos \theta}{qx} \\

E_y = \frac{2 * 9.1 * 10^{-31} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\

E_y = 3311.13 N/C\\E_y = 3.311 * 10^3 N/C

If the particle is a proton:

E_y = \frac{2 m_p u^2 sin \theta cos \theta}{qx} \\

E_y = \frac{2 * 1.67 * 10^{-27} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\

E_y = 6.08 * 10^6 N/C

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