Two particles with charges q1 = q2 = 7.00 nC are separated by a distance of 1.00 m. Suppose that q1 is at the point x = -0.500 m a) Make a diagram of the situation b) At what point on the line connecting the two charges, is the total electric field produced by both charges equal to zero? c) Suppose that a third electric charge q3 = 1.00 nC and mass m = 1 mg is placed at the point where the electric field is equal to zero and is given a small push. Determine the approximate value of the angular frequency of oscillation of q3 around this point, assuming small displacements around the equilibrium point. Hint: Use the binomial formula appropriately to make this approximation: (1 + x) n ≈ 1 + nx.

Two particles with charges q1 = q2 = 7.00 nC are separated by a distance of 1.00 m. Suppose that q1 is at the point x = -0.500 m a) Make a diagram of the situation b) At what point on the line connecting the two charges, is the total electric field produced by both charges equal to zero? c) Suppose that a third electric charge q3 = 1.00 nC and mass m = 1 mg is placed at the point where the electric field is equal to zero and is given a small push. Determine the approximate value of the angular frequency of oscillation of q3 around this point, assuming small displacements around the equilibrium point. Hint: Use the binomial formula appropriately to make this approximation: (1 + x) n ≈ 1 + nx.

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Question

Two particles with charges q1 = q2 = 7.00 nC are separated by a distance of 1.00 m. Suppose that q1 is at the point x = -0.500 m

a) Make a diagram of the situation

b) At what point on the line connecting the two charges, is the total electric field produced by both charges equal to zero?

c) Suppose that a third electric charge q3 = 1.00 nC and mass m = 1 mg is placed at the point where the electric field is equal to zero and is given a small push. Determine the approximate value of the angular frequency of oscillation of q3 around this point, assuming small displacements around the equilibrium point. Hint: Use the binomial formula appropriately to make this approximation: (1 + x) n ≈ 1 + nx.

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