An electron moving to the right at 8.5×10 5 m/senters a uniform electric field p

Question

An electron moving to the right at 8.5×105 m/senters a uniform electric field parallel to its direction of motion. If the electron is to be brought to rest in the space of 8.0 cm .

Part A

What direction is required for the electric field?

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Part B

What is the strength of the field?

Express your answer using two significant figures.

An electron moving to the right at 8.5×105 m/senters a uniform electric field parallel to its direction of motion. If the electron is to be brought to rest in the space of 8.0 cm .

Part A

What direction is required for the electric field?

What direction is required for the electric field? right up down left

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Part B

What is the strength of the field?

Express your answer using two significant figures.

Explanation / Answer

a) Remember that the direction of the electric at a point is the same direction as the force.

According to your question, if the electron is moving to the right , that means there is a heavy amount of charge

attracting it on the other end that is right .Electric field always moves with direction of force. Right direction.

b)

v=8.5*10^5m/s

x=8.0cm=0.08m

E=?

F=?

Q=1.60*10^-19

Since we have v, and x, at a uniform speed. u=0

we can use : v^2=u^2+2ax.

Now let look for acceleration (a) so we can latter find Force(F)

v^2=0+2ax

a=v^2/2x

a= (8.5*10^5)^2/(2*0.08)

a=7.225*10^11/0.16

a=4.515*10^12m/s^2

SO, back to Force (F)

F=ma

mass of an electron on a good looking day is : 9.11*10^-31kg

F=9.11*10^-31*4.515*10^12

F=4.113*10^-18N

SO, ELECTRIC FIELD IS FORCE PER CHARGE

E=F/Q

E=4.113*10^-18/1.60*10^-19

E=25.7 N/C

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