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Two large conducting platesX and Y. each having large surface area A (on one side), are placed parallel to each other as shown in figure (30-E7). The plate X ia given a charge Q whereas the other is neutral. Find (a) the surface charge dendity at the inner surface of the plates X, (b) the electric field at a point to the left of theplates, (c) the electric field at a point in between the plates and (d) the electric field at a point to the right of the plates. |
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Answer» Solution :(a) For the SURFACE charge density of a single plate. Let the surface charge density at both side be `sigma_1 and sigma_2`. Now ELECTRIC FIELD at both ends `= sigma_1/2 eplison_0 and sigma_2/2 eplison_0` Due to net balanced electric field on plate and `r_1 equal to r_2` `q_1= q_2 = Q/2 ` Net surface charge density = `Q/ 2A` (b) Electricfield to the left of the plates Since`r = Q/ 2A ` . Hence, electric field = `Q/2A(epsilon_0)` This must be directed towards left as 'X' is the charge plate. (c) and(d) here in both the cases the charged plate 'X' acts as the only SOURCE of electric field, with +ve in the inner side and'Y' attracts towards it with -ve in its innerside. So, for the middle portion E is towards right. (d) Similarly for extreme right the OUTER side of the 'Y' plate acts as +ve and hence it repels to the right with E = `Q/ 2A epsilon_0`. |
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