1) What is the magnitude of the force in newtons exerted on a charge of + 1.57 m
ID: 1572150 • Letter: 1
Question
1) What is the magnitude of the force in newtons exerted on a charge of + 1.57 microcoloumbs and mass 0.46 kilograms that is in an electric field of magnitude 8.81 meganewtons/coulomb? (no direction given)
2) An electric force is just that, a force, so you have to take Newton's Laws into account. What is the magnitude of the acceleration in metres per seconds squared if a charged object with a net charge of negative 1.9 millicoulombs and mass 7.8 grams is placed in an electric field of magnitude 402 newtons per coulomb?
3)A small charged particle of mass 2.95 milligrams and charge 8.99 microcoulombs is accelerated from rest through a voltage difference of 7,029 volts. What is the magnitude of the final velocity in metres per second?
Explanation / Answer
Force =qE
F = 1.57×10^-6 * 8.81*10^6
F = 13.83 N
2) F =qE
F= 1.9×10^-3×402 = 0.7638 N
F=ma
a= 0.7638/(7.8*10^-3)
a= 97.92 m/s^2
3)applying conservation of energy
KE i+PE i +w = KE f + PE f
0.5mv^2 = q * potential difference
V= sqrt(2qv/m)
V = sqrt(2*8.99*10^-6*7029/2.95*10^-3)
V =42.84 m/s
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.