electrostatic mc (physics)－vbd391t的部落格

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electrostatic mc (physics)

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1. Let T be the tension in the string hence, we have, in the horizontal direction, T.sin(theta) = qE, where q is the charge on the ball. In the vertical direction, T.cos(theta) = mg, where m is the mass of the ball and g is the acceeration due to gravity Dividing the two equations, tan(theta) = qE/mg i.e. E = (mg/q).tan(theta) Hence, E is proportional to tan(theta) as (mg/q) is a costant 2. Force of attraction Fp due to charge at P = k(-Q)q/(2a^2) where a is the length of the side of the square, and k is a constant hence, component of force along the QR direction = [-kQq/2a^2].cos(45) Force of repulsionFqalong QR due to charge at Q = kq^2/a^2 Since there is no net force along the QR direction, kq^2/a^2 + [-kQq/2a^2].cos(45) = 0 i.e. q = (Q/2).cos(45) q = square-root[2].(Q/4) The answer should be option B 3. Electirc force acting on the drop = (-1.6x10^-19) x (2000/0.01) N = -3.2x10^-14 N Since the drop is stationary, mg + (-3.2x10^-14) = 0 where m is the mass of the drop and g is the acceleration due to gravity hence, m = 3.26 x 10^-15 kg (take g = 9.81 m/s2) 4. Since the charge at X is +ve, the electric field it produces is going outward in direction from X to Y The charge at Y is -ve, electric field it produces is along direction from X to Y. It is then clear that the resultant electric field arising from the two charges is in the same direction, along the direction from X to Y. Potential at O due to charge 4 uC = k(4)/a where a is the distance from X to O, and k is a constant Potential at O due to charge at Y = k(-4)/a Hence, resultant pontential at O = 4k/a + (-4k/a) = 0

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