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Over [174]
3 years ago
5

If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.

Physics
1 answer:
Damm [24]3 years ago
5 0

Answer:

The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

Explanation:

As data is incomplete here, so by seeing the complete question from the search the data is

vx_0=1.1 x 10^6

ax=0 As acceleration is zero in the horizontal axis so

Equation of motion in horizontal direction is given as

s_x=v_x_0 t

t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s

Now for the vertical distance

vy_o=0

than the equation of motion becomes

s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2

Now using this acceleration the value of electric field is calculated as

E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

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Learn more about Pv diagrams : brainly.com/question/25401637

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