Two square, oppositely charged conducting plates measure 23 cm on each side. The plates are close together and parallel to each other. They each have a total charge of +4.6 nC and -4.6 nC, respectively.
A.)What is the electric field between the plates?
B.)What force is exerted on an electron located between the plates?
C.)What would be the electron's acceleration if it were released from rest?
I was understanding all of the other problems but I can't seem to come up with the correct answer for this one. Thanks for the help!
A.)What is the electric field between the plates?
B.)What force is exerted on an electron located between the plates?
C.)What would be the electron's acceleration if it were released from rest?
I was understanding all of the other problems but I can't seem to come up with the correct answer for this one. Thanks for the help!
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Since you don't know the separation of the plates you can't work out the capacitance and the voltage. So you have to use Gauss's Law.
A) Area of each plate = 0.23x0.23 = 0.0529m^2
Construct a Gaussian surface enclosing one plate. As the plates are close, the field is perpendicular to the plates and entirely from one late to the other, so
EA = Q/eo
E = 4.6x10^-9/(8.854x10^-12 x 0.0529) = 9821 = 9800V/m
B) F = eE = -1.6x10^-19x9821 = -1.57x10^-15N (negative sign means in the opposite direction to the field, ie. from -ve to +ve)
C) F = ma so a = F/m = 1.57x10^-15/9.11x10^-31 = 1.7x10^15m/s^2 in same direction as force.
A) Area of each plate = 0.23x0.23 = 0.0529m^2
Construct a Gaussian surface enclosing one plate. As the plates are close, the field is perpendicular to the plates and entirely from one late to the other, so
EA = Q/eo
E = 4.6x10^-9/(8.854x10^-12 x 0.0529) = 9821 = 9800V/m
B) F = eE = -1.6x10^-19x9821 = -1.57x10^-15N (negative sign means in the opposite direction to the field, ie. from -ve to +ve)
C) F = ma so a = F/m = 1.57x10^-15/9.11x10^-31 = 1.7x10^15m/s^2 in same direction as force.
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First calculate the voltage difference (potential difference). The electric field is the rate of change of V, or V/d where d = separation between the plates.
The force is qE where q = charge of an electron.
The acceleration is F/m where m = mass of electron.
How do you calculate V? This is a parallel plate capacitor. The capacitance is epsilon*A/d where A = area in square meters and d = separation in meters. epsilon = 8.854 x 10^-12 Farad/m
Plug all that in, get the capacitance, and use C = Q/V to solve for the voltage difference.
The force is qE where q = charge of an electron.
The acceleration is F/m where m = mass of electron.
How do you calculate V? This is a parallel plate capacitor. The capacitance is epsilon*A/d where A = area in square meters and d = separation in meters. epsilon = 8.854 x 10^-12 Farad/m
Plug all that in, get the capacitance, and use C = Q/V to solve for the voltage difference.