Please help !!! Solve the following 1.6 X 101 C) is released from rest in a unif
ID: 3308316 • Letter: P
Question
Please help !!! Solve the following 1.6 X 101 C) is released from rest in a uniform electrie field of magnitude 2.0 X 10 N/C. a. What is the magnitude of the force on the Ans. acceleration of the proton due to the electric field? the b. What is the magnitude of the Ans. of 2 X 10* kg is held suspended in air by an upward electric field 2. A charged oil drop with a mass of 300 N/C. a. What is the sign of the charge? b. Find the magnitude of the charge on the drop. Ans. Ans 3. The magnitude of the electrostatic force between two identical positive ions that are separated by 5 X 10 10 m is 3.7 X 109 N a. What is the charge on each ion? Ans. b. How many electrons are removed from each ion in-order to give it this charge? Ans.Explanation / Answer
Forc, F, on charge Q due to eleelect fiekf E is given by
F=QE
F=1.6*10^-19 * 2.0*10
F=3.2*10^-15 C.
Magnitude of acceleration can be found by
Force =mass * acceleration
3.2*10^-15=1.67*10^-27 *acceleration
Acceleration =3.2*10^-15/1.67*10^-27=1.92*10^12m/s².
2):the force on charged particle due to weight is downwards. There must be some upward force balancing it. Since electric field is positive, which means that charge is also POSITIVE as force on positive charge is in the direction of electric field.
b) : for equilibrium we have
Electrostatic force=gravitational force
QE=mg
Q*300=2*10 *10
Q=200*10³/300
Q=2*10³/3=0.666*10³=666 coloumbs.
3):let charge on each ion be q coloumb
Force=kqq/r where k=constant =9*10^9 and r=distance between two charges.
3.7*10-9=9*10^9*q²/5*10^-10
q²=3.7*10^-9*5*10^-10/9*10^9
q²=(3.7*5/9)*10^-28
q²=2.05*10^-28
q=1.43*10^-14 coloumb.
b) ;charge on one electron=-1.6*10^-19
Total charge=1.43*10^-14
Charge=no.of electrons* charge on one electron
1.43*10^-14=n*1.6*10^-19
n=0.89375*10^5
n=89375 electrons should be removed to get the above amount of charge.