Two parallel plates having charges of equal magnitude but opposite sign are sepa

Question

Two parallel plates having charges of equal magnitude but opposite sign are separated by 17.0 cm. Each plate has a surface charge density of 30.0 nC/m2. A proton is released from rest at the positive plate.

(a) Determine the potential difference between the plates.
____?_____ V

(b) Determine the kinetic energy of the proton when it reaches the negative plate.
____?______ J

(c) Determine the speed of the proton just before it strikes the negative plate.
_____?_____ km/s

(d) Determine the acceleration of the proton.
____?_______ m/s2 (towards the negative plate)

(e) Determine the force on the proton
_____?______N (towards the negative plate.)

(f) From the force, find the magnitude of the electric field.
____?________kN/C

Show that it is equal to that electric field found from the charge densities on the plates.

Explanation / Answer

here

part A:

Electric field E = sigma/eo

E = 30 e -9/8.85 e -12

E = 3389.83 N/C

Potential V = Ed

V = 3389.83* 0.17

V = 576.27 Volts

----------------------------------

b. KE = eV

KE = 1.6 e-19 * 576.27

KE = 9.22032e-17 Joules

-------------------------

0.5 mv^2 = 9.22 e -17

v = sqrt(2 *9.22 e-17/1.67 e -27)

v = 3.322 e 5 m/s

-----------------------------------------------

use ma = Eq

a = 3389.83* 1.6 e -19/(1.67 e-27)

a = 3.24 e 11 m/s^2

----------------------------------------

Force F = ma

F = 1.67 e -27 * 3.24 e 11

F = 5.41 e -16 N

---------------------------------------
force F = Eq

F = 3389.83 * 1.6 e-19

F = 5.423 *10^-16 N

------------------------------------------------

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.