3. What is the magnitude of the electric field at a point 0.45 cm away from a po
ID: 1438024 • Letter: 3
Question
3. What is the magnitude of the electric field at a point 0.45 cm away from a point charge of +4.5 pC ?
4.An electron is acted on by an electric force of 5.2×1014 N . What is the magnitude of the electric field at the electron's location?
5.An electron is acted on by an electric force of 5.2×1014 N . What is the magnitude of the electric field at the electron's location?
6.Calculate the voltage required to accelerate a beam of electrons initially at rest if they have a kinetic energy of 3.5 keV . Calculate their speed if they have a kinetic energy of 3.5 keV .
7.Calculate the voltage required to accelerate a beam of electrons initially at rest if they have a kinetic energy of 5.5×1016 J .Calculate their speed if they have a kinetic energy of 5.5×1016 J .
Explanation / Answer
3) This is just putting numbers into the standard formula.
Make sure you get the units correct:
0.45cm = 0.0045m
4.5pC = 4.5x10¹²C
E = kq/r²
= (9x10) x (4.5x10¹²) / 0.0045²
=2000N/C to 3 significant figures
4) I assume that the force is in N.
Electric field intensity is the force experienced by unit charge.
Electric field at the location of the electron = E = F/e
E = 5.2x10^-14/1.602x10^-19 = 3.5x10^5 N/C
5) Same to question (4)
6) the energy is now 3.5keV, e.g. 3500eV
Remember that 1 eV = 1.602 * 10-19 J
Also remember that the mass of a proton = 1.6726 * 10-27 kg
KE = 3500 * (1.602 * 10-19 J / eV)
KE = 5.607 * 10-16
KE = 0.5 * m * v2
5.607 * 10-16 = 0.5 * (1.6726 * 10-27) * v2
v =7.95 * 10^5 m/s
7.95 * 10^5 m/s
7)
the energy is now 5.5*10^-16
Remember that 1 eV = 1.602 * 10-19 J
Also remember that the mass of a proton = 1.6726 * 10-27 kg
KE =5.5*10^-16 * (1.602 * 10-19 J / eV)
KE = 8.811*10^-35
KE = 0.5 * m * v2
8.811 * 10-35 = 0.5 * (1.6726 * 10-27) * v2
v =3.152 * 10^-4 m/s
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