http://www.webassign.net/bauerphys1/24-p-031.gif picture A capacitor consists of
ID: 2026330 • Letter: H
Question
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A capacitor consists of two parallel plates, but one of them can move relative to the other as shown in the figure. Air fills the space between the plates, and the capacitance is 39.0 pF when the separation between plates is d = 0.480 cm.
(a) A battery with potential difference V = 9.0 V is connected to the plates. What is the charge distribution, s, on the left plate?
What are the capacitance, C', and charge distribution, s', when d is changed to 0.240 cm?
C' =
s' =
(b) With d = 0.480 cm, the battery is disconnected from the plates. The plates are then moved so that d = 0.240 cm. What is the potential difference V', between the plates?
Explanation / Answer
Given data: Capacitance, C = 39 pF = (39 pF) (10-12 F / 1 pF) = 39 *10-12 F Distance between the plates, d = 0.480 cm = (0.480 cm) (10-2 m / 1 cm) = 0.0048 m Potential difference, = 9.0 V ------------------------------------------------------------------------ Solution: (a) The capacitance of a parallel plate capacitor is, C = 0 A /d Charge on a capacitor is, Q = C V Q =( 0 A /d) V Q / A = 0 V /d ...... (1) But charge distribution, s = Q / A Now, equation (1) becomes s = 0 V /d ...... (2) = (8.85 *10-12 C2 / N.m2) (9.0 V / 0.0048 m) = 1.65 *10-8 C / m2 ---------------------------------------------------------------------------------- If d ' = 0.240 cm = (0.240 cm) (10-2 m / 1 cm) = 0.0024 m Since C = 0 A /d, C / C' = d' / d C' = ( 39 *10-12 F) (0.0048 m/0.0024 m) = 78 *10-12 F or 78 pF ------------------------------------------------------------------------------------ Charge distribution, s' = s (d / d') [From equation (2)] = (1.65 * 10-8 C/m2) (0.0048 m/0.0024 m) = 3.3* 10-8 C/m2 ----------------------------------------------------------------------------------------- (b) d = 0.0048 m is changed to d ' = 0.0024 m, then the charge on the plates can be given using the relation Q = C V = (0 A /d) V Q ' / Q = d / d' Q' = Q (0.0048 m/0.0024 m) Q' = 2Q Now, Q = C V Q' / Q = V ' / V V' = V (2) = 9.0 V (2) = 18 V = (0.240 cm) (10-2 m / 1 cm) = 0.0024 m Since C = 0 A /d, C / C' = d' / d C' = ( 39 *10-12 F) (0.0048 m/0.0024 m) = 78 *10-12 F or 78 pF ------------------------------------------------------------------------------------ Charge distribution, s' = s (d / d') [From equation (2)] = (1.65 * 10-8 C/m2) (0.0048 m/0.0024 m) = 3.3* 10-8 C/m2 ------------------------------------------------------------------------------------ Charge distribution, s' = s (d / d') [From equation (2)] = (1.65 * 10-8 C/m2) (0.0048 m/0.0024 m) = 3.3* 10-8 C/m2 ----------------------------------------------------------------------------------------- (b) d = 0.0048 m is changed to d ' = 0.0024 m, then the charge on the plates can be given using the relation Q = C V = (0 A /d) V Q ' / Q = d / d' Q' = Q (0.0048 m/0.0024 m) Q' = 2Q Now, Q = C V Q' / Q = V ' / V V' = V (2) = 9.0 V (2) = 18 VRelated Questions
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