Question

An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is = 2.0 × 10-7 C/m2, and the plates are separated by a distance of 1.9 × 10-2 m. How fast is the electron moving just before it reaches the positive plate? ty for any help :)

Two charges are located on the x axis: q1 +5.5 1C at x1 = +5.4 cm, and 92 +5.5 LC at x2 =-5.4 cm. Two other charges are located on the y axis: q3 +6.4 cm. Find (a) the magnitude and (b) the direction of the net electric field at the origin. +3.5 1C at y3 +4.8 cm, and q- 10 LC at w 42-_1-+x (a) Number Units (b) The net electric field points

Answer 1

The two charges on the X axis are equidistant from the origin and also are identical. Hence their electric field at origin will be equal in magnitude and opposite in direction and will cancel out. Hence the electric field at origin is only due to Q3 and Q4.

origin. 1) a) q, and are are identical and equidistant from So their fields cancel out at the origin. So field is only due to 93 and 94 Enet = Ezt Eq - = -kousl ŷ + klaa) ŷ (0-048%)” (0-064w) 0-06+ YA 10.0484 = -K3.5 uc) î + kllous) 0.048m) (-064m) = 8:3X106 N/c So, magnitude is [8.38109N/ direction » toward ty axis or upward 2 E = 6 2x1077 C/m2 = 2:26X 10+ NIC Co acelne, a = Em = 9€ - LE - (.6X20993)2.26*10*nly) 9.11% 20-34 1g = 3.97X 1015 m/s2 v = ² + zas = 0+ 213.97X 1025m/s2) (1.9x102m) = 1.23x107) m/s

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This concludes the answers. If there is any mistake or omission, let me know in the comments immediately and I will fix it....