Question

A small metal sphere, carrying a net charge of q1 = -2.70 μC , is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2 = -7.50 μC and mass 1.70 g , is projected toward q1. When the two spheres are 0.800 m apart, q2 is moving toward q1 with speed 22.0 m/s (Figure 1). Assume that the two spheres can be treated as point charges.

What is the speed of q2 when the spheres are 0.400 m apart?

How close does q2 get to q1?

Answer 1

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Answer 2

To solve this problem, we can apply the principle of conservation of energy. The initial kinetic energy of sphere q2 is equal to the sum of the final kinetic energy and potential energy when the spheres are at a distance of 0.800 m apart. We can use this information to find the speed of q2 when the spheres are 0.400 m apart.

Let's denote the initial kinetic energy of q2 as K1, the final kinetic energy as K2, and the potential energy as U.

The initial kinetic energy of q2 is given by: K1 = (1/2) * m * v1^2

Where m is the mass of q2 and v1 is the initial speed of q2 (22.0 m/s).

The potential energy at a distance of 0.800 m apart is given by: U1 = k * (q1 * q2) / r1

Where k is the electrostatic constant (8.99 x 10^9 N*m^2/C^2), q1 is the charge of q1, q2 is the charge of q2, and r1 is the initial distance between the spheres (0.800 m).

The final kinetic energy of q2 when the spheres are 0.400 m apart is given by: K2 = (1/2) * m * v2^2

Where v2 is the final speed of q2.

The potential energy at a distance of 0.400 m apart is given by: U2 = k * (q1 * q2) / r2

Where r2 is the final distance between the spheres (0.400 m).

According to the principle of conservation of energy, the initial kinetic energy plus the initial potential energy is equal to the final kinetic energy plus the final potential energy: K1 + U1 = K2 + U2

Substituting the expressions for K1, U1, K2, and U2, we can solve for v2:

(1/2) * m * v1^2 + k * (q1 * q2) / r1 = (1/2) * m * v2^2 + k * (q1 * q2) / r2

Simplifying the equation, we get:

(1/2) * m * v1^2 - (1/2) * m * v2^2 = k * (q1 * q2) * (1/r2 - 1/r1)

Now we can solve for v2:

(1/2) * m * (v1^2 - v2^2) = k * (q1 * q2) * (1/r2 - 1/r1)

v2^2 = v1^2 - (2 * k * (q1 * q2) * (1/r2 - 1/r1)) / m

Finally, we can calculate v2 using the given values of q1, q2, m, v1, r1, and r2:

q1 = -2.70 μC q2 = -7.50 μC m = 1.70 g = 0.0017 kg v1 = 22.0 m/s r1 = 0.800 m r2 = 0.400 m

Substituting these values into the equation, we can solve for v2:

v2^2 = (22.0 m/s)^2 - (2 * (8.99 x 10^9 N*m^2/C^2) * (-2.70 x 10^-6 C) * (-7.50 x 10^-6 C) * (1/0.400 m - 1/0.800 m)) / 0.0017 kg

Simplifying the equation, we find:

v2^2 ≈ 485.88 m^2/s^2

Taking the square root of both sides, we get:

v2 ≈ 22.04 m/s

Therefore, the speed of q2 when the spheres are 0.400 m apart is approximately 22.04 m/s.

To determine how close q2 gets to q1, we can find the distance between the spheres at the closest approach. At the closest approach, the potential energy becomes zero. So we can set U2 = 0 and solve for r2:

0 = k * (q1 * q2) / r2

Simplifying the equation, we find:

r2 = k * (q1 * q2) / 0

Since the denominator is zero, we have an indeterminate form. This indicates that the spheres collide or come into contact with each other at the closest approach.