GOAL Calculate fundamental physical properties of a parallel- plate capacitor PR

Question

GOAL Calculate fundamental physical properties of a parallel- plate capacitor PROBLEM A parallel-plate capacitor has an area A = 2.00. 10-4 m2 and a plate separation d = 1.00-10.3 m. (a) Find its capacitance. (b) How much charge is on the positive plate if the capacitor is connected to a 3.00-V battery? Calculate (c) the charge density on the positive plate, assuming the density is uniform, and (d) the magnitude of the electric field between the plates. STRATEGY Parts (a) and (b) can be solved by substituting into the basic equations for capacitance. In part (c) use the definition of charge density, and in part (d) use the fact that the voltage difference equals the electric field times the distance SOLUTION (A) Find the capacitance Substitute values into the deinition of (8a5 x11.09 x 109 m C-1.77 101.77 p (B) Find the charge on the positive plate after the capacitor is connected to a 3.00-V battery. Substitute into the capacitance charge c"erar g"car " (1.77 × 10.12 'C3.00 V) equation to ind the charpe -5.11 x 1012c (C) Calculate the charge density on the positive plate. Charge density is charge divided by 4 2.00 x 1m (D) Calculate the magnitude of the electric field between the plates. 3.00V 1.00x 10m LEARN MORE REMARKS The answer to part (d) could also have been obtained from the electric field derived for a parallel plate capacitor E = /eo. QUESTION How do the following change if the distance between the plates is doubled? (Select all that apply.) O The electric field between the plates is halved. O The electric field between the plates is doubled. O The charge is halved.

Explanation / Answer

Electric field E = Q/(A*eo)

potential difference dV = E*d

capacitance C = eo*A/d

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The electric field remains the same

The capacitance is halved

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(a)

capacitance = eo*A/d = 8.85*10^-12*2.5*10^-4/(1.1*10^-3) = 2*10^-12 F

(b)

charge Q = C*v = 2*10^-12*3 = 6*10-12 C

(c)

charge density = Q/A = (6*10^-12)/(2.5*10^-4) = 2.4*10^-8 C/m^2

(d)

E = V/d = 3/(1.1*10^-3) = 2.73*10^3 V/m

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(a)

C = eo*A/d

d = eo*A/C

d = 8.85*10^-12*3.3*10^-4/(1.1*10^-12)

d = 2.655*10^-3 m

(b)

charge density = Q/A = (4.8*10^-12)/(3.3*10^-4) = 1.45*10^-8 C/m^2

(c)

Electrc field E = Q/(A*eo) = 4.4*10^-12/(3.3*10^-4*8.85*10^-12) = 1.506*10^3 N/C

(d)

V = Q/C

V = 4.8*10^-12/(1.1*10^-12)

V = 4.36 v

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