Problem 11: An electron with an initial speed of 29 × 105 m/s is fired in the same direction as a uniform electric field of 80 N/C. How far does the electron travel before brought to rest and turned back?
Solution
Since electron has negative charge, therefore, it will be retarded in the given conditions. Using the equation F = q0E (Force on the electron in the electric field) and then equating with F = ma. Remember that charge on an electron is 1.6 × 10-19 C.
Now mass of an electron is 9.1 × 10-31 Kg, therefore,
Here the negative sign accounts for the direction of the acceleration (retardation).
Now to find the distance the electron covers before coming to rest (vf = 0) can be calculated by using the equation of motion, 2as = vf2 – vi2. Put the values,
How you take the force negative when u put it in equation f=ma
Aly
Yes, this is important question.
This is clear that the electron experiences a retarding force. We want to know about the magnitude of this force which is determined by the formula q0E. Once the magnitude of this force is known the next step would be to find how long will it travel in the field. So, we apply equation f = ma and proceed as in the text above. So, the first equation f = q0E gives us the magnitude of the force and as the motion of electron and the electric field are in the same direction, the electron will be retarded and the direction of the force will be opposite to that of the direction of motion of the electron. Therefore, we take the force to be negative, stopping the moving electron.
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