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IceJOKER [234]
3 years ago
14

A proton, initially traveling in the +x-direction with a speed of 5.05×10^5 m/s , enters a uniform electric field directed verti

cally upward. After traveling in this field for 3.90×10^−7 s , the proton’s velocity is directed 45° above the +x-axis. What is the strength of the electric field?
Physics
1 answer:
Korvikt [17]3 years ago
3 0

Answer:

The strength of the electric field is 1.35\times10^{4}\ N/C.

Explanation:

Given that,

Speed v= 5.05\times10^{5}\ m/s

Time t= 3.90\times10^{-7}\ s

Angle = 45°

We need to calculate the acceleration

Using equation of motion

v = u+at

5.05\times10^{5}=0+a\times3.90\times10^{-7}

a =\dfrac{5.05\times10^{5}}{3.90\times10^{-7}}

a=1.29\times10^{12}\ m/s^2

We need to calculate the strength of the electric field

Using relation of newton's second law and electric force

F= ma=qE

ma = qE

E=\dfrac{ma}{q}

Put the value into the formula

E=\dfrac{1.67\times10^{-27}\times1.29\times10^{12}}{1.6\times10^{-19}}

E=1.35\times10^{4}\ N/C

Hence, The strength of the electric field is 1.35\times10^{4}\ N/C.

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