Question

Problem 1 A parallel plate capacitor has bottom plate on the x-y plane. The sepa- ration between the plates is d. The plates each have width along the y-axis w, and length along the r-axis, l. Between the plates there is a dielectric material of the same dimensions. It has dielectric constant e,. We move the dielectric material rt so that it sticks out of one end of the capacitor and leaves a gap between the plates at the other end. Show that the capacitance is given by: (e,l- Xet) C =

Answer 1

solution :- llale-/ -dielefic - L plalt.2 (fig.1) biven, a parallel plale capacitor placed in xy piace. The separation b/w the places = The area of the plates = exw a also the dielupic of same dimension in introduced to them. Now, The distance a i . dietafic (Ex) has moved a cesta col c L-2 Now By the displaunens of the dielufsic two medium of capacitance code generated Now, colc are in parallel combination, since potential difference remains same.

now equivalenf copacilence Ceg = 9+2 = cota anco + Alote A₂totr cl ...A,= Kw Е. + (2-2)w to ty l Az = (1-2)W d browo [ 2+(1-2)67] - Favo [ 676 + (1-6x)2] trou ( 418-(6-1) z] [:de: 6703 seus [4,- 862] o com Nibe for any other con query feel free to comment ... like .....