find the magnitude and direction of the electric field.
the initial velocity of an electron moving along the x axis in an uniform electric field is 3x10^6ms^-1.if the electron stop after distance of 45cm,find the magnitude and direction of the electric field.
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The electric force that acts on the electron is
`F_e = -e*E` (1)
This force is giving the electron an acceleration of
`a = F_e/m = -(e/m)*E`
The equation of motion of the electron in the electric field is
`V^2 = V_0^2 +2*a*s`
Here `V =0 m/s` , `V_0 = 3*10^6 m/s` , `s =45 cm =0.45 m`
`0 =V0^2 -2*(e/m)*E`
`E = (V_0^2/2)*(m/e) =((9*10^12)/2)*((9.1*10^-31)/(1.6*10^-19)) =25.59 V/m`
Since the electron has an initial positive speeds it means it is moving toward the right (if the x axis is from left to right). To stop the electron the electric force (1) needs to oppose the initial speed of the electron, hence it is to the left. Because the charge of the electron (`-e`) is negative, the electric field that generates this force is directed from left to the right (positive direction of x axis).
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