A capacitance is made of two parallel conducting plates as shown in the figure and the distance between the plates is 'd'. (i) write the value of capacitance as a function of all the possible dimensions from L,W, h and d (ii) if a dielectric slab of thickness 't' and area of cross-section A (t<d), and relative permittivity Er is inserted in between the plates, what will be the value of capacitance (i) If a voltage V is applied between the plates, find the voltage across various sections (iv) If the dielectric slab is of length L/2 and width 'W and relative permittivity Er and is inserted in between the plates from right end, what is the value of capacitance.
i) Capacitance for parallel plate capacitor is given as ,
"c=\\frac{E_0A}{d}, A=WL"
or "c=\\frac{E_0WL}{d}(F)"
V"=2V_1+V\n_3"
"=2V_1+\\frac{c_1}{c_3}v_1"
"=(2+\\frac{2E_0WL}{d-t}*\\frac{t}{E_0E_rWL})V_1" )
"=V_1=\\frac{E_r(d-t)}{2[t+E_r(d-t)]}"
"V_3=\\frac{2t}{E(d-t)}.\\frac{E_r(d-t)}{2[t+E_r(d-t)]}"
Thus, "V_1=V_2=\\frac{E_r(d-t)V}{2(t+E_r(d-t)},V_3=\\frac{tV}{t+E_r(d-t)}"
ii) "c_1=c_2=\\frac{E_0(WL)}{\\frac{d-t}{2}}"
"c_1=c_2=\\frac{2E_0WL}{d-t}"
"c_3=\\frac{E_0E_rWL}{t}"
"\\frac{1}{c_9}=\\frac{1}{c_1}+\\frac{1}{c_2}+\\frac{1}{c_3}"
"\\frac{d-t}{2E_0E_rWL}+\\frac{d-t}{2E_0WL}+\\frac{t}{E_0E_rWL}"
"\\frac{1}{C_q}=\\frac{d-t}{2E_0WL}+\\frac{t}{E_0E_rWL}"
"C_q=\\frac{E_0E_rWL}{t+E_r(d-t)}"
iii)V"=V_1+V_2+V_3"
"Q=C_2V"
"\\therefore C_1=C_2 \\space thus \\space V_1=V_2"
iv) "C_1=\\frac{E_0WL}{2d}"
"C_2=\\frac{E_0WL}{d-t}"
"C_3=\\frac{E_0E_rWL}{2t}"
"=C_2" & "C_3" are in series
"C_{eq2}=\\frac{C_2C_3}{C_2+C_3}=\\frac{E_0E_rWL}{2(t+E_r(d-t)}"
"C_{eq2}=C_1+C_{eq1}"
"=\\frac{E_0WL}{2d}+\\frac{E_0E_rWL}{2(t+E_r(d-t)}"
"C_{eq}=\\frac{E_0WL}{2}+[\\frac{1}{d}+\\frac{E_r}{t+E_r(d-T)}]"
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Please answer Question number #199623 in Electrical Engineering as soon as possible....by tomorrow please
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