1. An electron in a certain region of space experiences a force of 4.8 10-15N ×

Question

1. An electron in a certain region of space experiences a force of 4.8 10-15N × due North from the electric
field there. Find the magnitude and direction of the electric field.
2. What is the magnitude and direction of the electric force on an ion Mg++placed at the same location as
the previous problem?
3. What is the acceleration of the ion in the previous problem, given that its mass number is 24?
4. A point charge of 20nC is placed on the x-axis at the point x = 2. Find the x-component of the electric
field x E at x = 4 and x = -1where the coordinates are in meters. Sketch x E as a function of x .
5. Repeat the previous problem if the point charge is - 20nC

Explanation / Answer

An electron in a certain region of space experiences a force of F =4.8* 10^-15N charge of the electron is e =1.6*10^-19 C the electrical force on the electron due to electric feild is F =e*E therefore electric feild E =F/e                                      =4.8*10^-15 N /1.6*10^-19 C                                      =3*10^4 N/C the direction of the electric feild is : towards South direction _______________________________________________________________________________ 2. the given Mg++ ion has charge q = 2 proton charge                                                  =2*1.6*10^-19 C                                                  =3.2*10^-19 C from the above problem electric feild is E =3*10^4 N/C therefore electric force on Mg++ is F = q*E                                                          =3.2*10^-19C *3*10^4 N/C                                                          =9.6*10^-15 N direction : South ______________________________________________________________________________ electric force on the Mg++ is F =9.6*10^-15 N from Newton's laws F =ma mass of the Mg++ is m = 24*mass of the proton                                    =24*1.6*10^-27 kg                                    =0.4*10^-25 kg acceleration of the ion Mg++ is a =F/m                                                    =9.6*10^-15 N/0.4*10^-25                                                    =2.4*10^11 m/s^2 ____________________________________________________________________________ 4. electric feild E =1/40(q/x^2)...........(1) x =2 m charge q =20nC =20*10^-6 C 1/40=9*10^9 Nm^2/C^2 substtuting these values and in equation (1) and find out the x-component of electric feild repeat this for x =4 m and x =-1means x =1 m ____________________________________________________________________________ repeat problem 4 for q =-20 nC =-20*10^-6 C

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