Question

Soalan 1 Question 1 (a) Pertimbangkan kapasitor silinder dua konduktor sepaksi dengan radius dalaman a dan radius luaran b (b>a). Ruang di antara konduktor dipenuhi dengan dielektrik yang seragam dengan ketelusan e,. Andaikan bahawa konduktor 1 dan 2 membawa cas +Q dan-Q, masing-masing yang diedarkan secara seragam ke atasnya. Consider a cylindrical capacitor of two coaxial conductors of inner radius a and outer radius b (b > a). The space between the conductors is filled with a homogeneous dielectric with permittivity ep. Assume that conductor 1 and 2 carry charge +Q and-Q respectively which are uniformly distributed on them. i. Terbitkan keamatan medan elektrik E dan beza keupayaan V pada titik r di mana (a<r< b). Derive the electric field intensity E and electric potential V at a pointr where (a<r>b). (5 markah / marks) ii.Dari (i), jika medium dielektrik adalah udara, selesaikan kapasitan per meter kapasitor silinder apabila a 0.5 cm dan b = 2.0 cm From (i), if the dielectric medium is air, solve the capacitance per meter of the cylindrical capacitor when a 0.5 cm and b 2.0 cm (4 markah / marks) ii. Dari (i), jika ruang di antara plat diisi dengan dielektrik tidak sekata dengan &r (5+r)/r, di mana r adalah dalam sentimeter, sila nilai kapasitan baru per meter kapasitor silinder. From (ii), if the space between the plates is filled with an inhomogeneous dielectric with e = (5 +r)/r, where r is in centimetres, evaluate the new capacitance per meter of the cylindrical capacitor (7 markah/marks) iv. Banding dan terangkan kapasitan per meter di dalam (ii) and (i) Compare and explain the capacitance per meter in (ii) and (ii). (4 markah/marks)

Answer 1

For Q = lambda l where l is length of cable at p E = 2/2 pi E r [contourintegral E^bar middot ds = Q/E] E = (Q/l)/2 pi E r E = Q/2 pi E rl V = integral^r_a E^bar middot dr = integral^r_a (Q/l)/2 pi E r dr = (Ql) 1/2 pi E ln [r/a] \r\n V for r = b = a/l(2 pi E) ln (2/0.5) for l = 1 m V = Q/2 pi E ln (4) C = Q/V = Q/Q/2 pi E ln (4) = 2 pi E/ln 4 = 2 times 2.851 times 10^-12/ln 4 = 40.1159 times 10^-12 F E = Q/l(2 pi E_r E_0 r) V = integral E middot dr = integral^b_a Q/2 pi l E_r E_0 r dr \r\n l = 1 m V = integral^b_a Q/2 pi(5 + r/r) E_0 r dr = Q/2 pi E_0 ln(5 + r)^b_a = Q/2 pi E_0 ln(5 + r)^2_0.5