Solved Exercise/Numericals - Electric Field Intensity (E) & Force (F) - 1.
Q.1 Point charges Q 1 = 5 μC and Q 2 = -4 μC are placed at (3, 2, 1) and (-4, 0, 6) respectively. Determine the force on Q 1 . Ans: Force on Q 1 due to Q 2 is given as : = -5.74 a x – 1.642 a y + 4.104 a z mN Q. 2 Five Identical 15 μC point charges are located at the center and corners of a square defined by -1 < x, y > 1 and z = 0. Find the force on the 10 μC point charge at (0, 0, 2). Therefore, F / 1.35 = [(0, 0, 2) / 8] + [(-1, -1, 2) / 6 3/2 ] + [(1, -1, 2) / 6 3/2 ] + [(-1, 1, 2) / 6 3/2 ] + [(1, 1, 2) / 6 3/2 ] → F = 1.35 [(0.25 + (8 / 6 3/2 )] a z = 1.072a z N Q.3 Point Charges Q 1 and Q 2 are respectively located at (4, 0, -3) and (2, 0, 1). If Q 2 = 4nC, find Q 1 such that a) The E at (5, 0, 6) has no z component. b) The force on a test charge at (5, 0, 6) has no x component. a...