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ch25 tủ tài liệu training

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Nội dung chi tiết: ch25 tủ tài liệu training

ch25 tủ tài liệu training

I.Charge flows until the potential difference across the capacitor is the same as the potential difference across the batter)'. The charge on the capa

ch25 tủ tài liệu trainingacitor is then q = cr, and this is the same as the total charge that has passed through the batter)’. Thus,q = (25x10* F)( 120 V) = 3.0 X I O’3 c.2(a)

The capacitance of the system isc = jf_= 70pCAK 20V= 35 pF.(b)The capacitance is independent of q\ it is still 3.5 pF.(c)rhe potential difference bec ch25 tủ tài liệu training

omes200pCc 35 pF= 57V.3(a) The capacitance of a parallel-plate capacitor is given by c - ZfiAld, where A is the area of each plate and d is the plate

ch25 tủ tài liệu training

separation. Since the plates are circular, the plate area is A JiR2, where R is the radius of a plate. Thus,?.JĩR- (8.85xlO-nF/m>(8.2x1O-’m) ,_ In_c=1

I.Charge flows until the potential difference across the capacitor is the same as the potential difference across the batter)'. The charge on the capa

ch25 tủ tài liệu trainingference across the plates. Thus.4 = (1.44 X 10'10F)( 120 V) = 1.73 X 10“* c= 17.3 nC.4We use c - Acofd.(a) Thus,A? l.OOnr 8.85xl0-|2-£yư = ^ = ------

-----------------— = 8.85 X IO-'2 m.c1.00 F(b) Since ch25 tủ tài liệu training

ming conservation of volume, we find the radius of the combined spheres, then use c - to find the capacitance. When the drops combine, the volume is d

ch25 tủ tài liệu training

oubled. Il is then v= 2(4n/3)/?'. The new radius R' is given by=> R' = 2ViR.3’3The new capacitance isC=An^R'=4K£O2Ự3 R = 5.04^/?.With R = 2.00 mm. we

I.Charge flows until the potential difference across the capacitor is the same as the potential difference across the batter)'. The charge on the capa

ch25 tủ tài liệu traininguired be A. Then c - ib.4/(/> - «). or

I.Charge flows until the potential difference across the capacitor is the same as the potential difference across the batter)'. The charge on the capa

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