Two parallel, infinite, nonconducting plates are 4.53 cm apart and have charge distributions of +2.25 μC/m2+2.25 μC/m2 and –2.25 μC/m2–2.25 μC/m2 . (a) What is the magnitude of the force on an electron in the space between the plates? (You may enter your calculation using scientific notation.) N (b) What is the magnitude of the force on an electron located outside the two plates near the surface of one of the two plates? N
Question
Two parallel, infinite, nonconducting plates are 4.53 cm apart and have charge distributions of +2.25 μC/m2+2.25 μC/m2 and –2.25 μC/m2–2.25 μC/m2 . (a) What is the magnitude of the force on an electron in the space between the plates? (You may enter your calculation using scientific notation.) N (b) What is the magnitude of the force on an electron located outside the two plates near the surface of one of the two plates? N
Solution
To solve this problem, we first need to find the electric field between the plates. The electric field E created by a charged plate is given by the equation E = σ/2ε0, where σ is the surface charge density and ε0 is the permittivity of free space (8.85 x 10^-12 C^2/N*m^2).
(a) The total electric field between the plates is the sum of the fields created by each plate. Since the plates have equal and opposite charges, their fields add up:
E = σ/2ε0 + σ/2ε0 = σ/ε0
Substituting the given value for σ (2.25 x 10^-6 C/m^2), we find:
E = 2.25 x 10^-6 C/m^2 / 8.85 x 10^-12 C^2/N*m^2 = 2.54 x 10^5 N/C
The force F on an electron in this field is given by F = qE, where q is the charge of the electron (-1.6 x 10^-19 C). The negative sign indicates that the force is in the opposite direction of the field.
F = -1.6 x 10^-19 C * 2.54 x 10^5 N/C = -4.06 x 10^-14 N
The magnitude of this force is 4.06 x 10^-14 N.
(b) Outside the plates, the fields created by each plate cancel each other out, so the total electric field is zero. Therefore, the force on an electron outside the plates is also zero.
Similar Questions
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