of E = 34.97 MV/m is desired between two parallel plates, each of area 4.1 cm2 and separated by 14.19 mm of air. What charge must be on each plate? Express your answer in nC. Note: For this problem, use ε0 = 8.8541878x10-12 F/m
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- Three charges are placed along the x-axis: q1 = 3.00 µC at X1 = -20.0 cm, q2 = -5.00 µC at x2 = 10.0 cm, and q3 = 6.00 µC at x3 = 25.00 cm. Determine the magnitude of the electric field at the origin. O 4 69x106 N/C O 2 96x106 N/C O 6 04x106 N/C O 4.31x106 N/CA test charge of 13 μC is at a point P where the electric field due to other charges is directed to the right and has a magnitude of 4 x 106 N/C. If the test charge is replaced with a charge of -3 μC, the electric field at P is2 narrow, flat metal plates are positioned vertically, 20.00 cm. The first plate has a positive charge with charge density σ=+630.0 mC/m2 and a second plate has an equal but opposite negative charge with charge density σ=-6300.0 mC/m2 . There are also two narrow, flat metal plates positioned horizontally, 30.00 cm apart, with the top plate given a negative charge, and the bottom plate given an equal but opposite positive charge, such that the electric potential of the bottom plate is 5.00 V higher than the top plate. A small sphere with a mass of m =64.35 g, and a charge of q =22.00 mC is attached to a narrow, stiff, massless, insulating rod with a length of L= 8.00 cm, which is pivoted at point O, which is 2.000 cm from the second plate. The sphere/rod unit is angled at 5 degrees with horizontal and released from rest. Will the sphere/rod ever hit an angle of 0 degrees with the horizontal? If so, how long will it take to reach that point?
- The electric field just above the surface of the charged drum of a photocopying machine has a magnitude E of 3.0 × 105 N/C. What is the surface charge density on the drum, assuming that the drum is a conductor? Number UnitsA single isolated, large conducting plate has acharge per unit area σ on its surface. Because the plate is a conductor, the electric field at its surface is perpendicular to the surface and has magnitude E = σ/εo a.The field from a large, uniformly charged sheet with charge per unit area σ has magnitude E = σ/2εo. Why is there a difference? b.Regard the charge distribution on the conducting plate as two sheets of charge (one on each surface), each with charge per unit area σ. Find the electric field inside and outside the plate.18. 6.0 m and total charge Q = 120 µC distributed uniformly along the length 8. Figure 5 shows a thin rod of length L of the rod. By direct integration, find the magnitude of the electric field at the point P in the figure. (s) Integral Figure 5 Call on X-axis no pat L/3 2 m 120ML. L L/2 3n a