Question
Asked Sep 21, 2019
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A small, stationary sphere carries a net charge Q. You perform the following experiment to measure Q: From a large distance you fire a small particle with mass 4×10^−4kgand charge 5×10^−8C directly at the center of the sphere. The apparatus you are using measures the particle's speed v
as a function of the distance x from the sphere. The sphere's mass is much greater than the mass of the projectile particle, so you assume that the sphere remains at rest. All of the measured values of x are much larger than the radius of either object, so you treat both objects as point particles. You plot your data on a graph of v2 versus (1/x)(Figure 1). The straight-line v2=400 m^2/s^2−[(15.75 m^3/s^2)/x] gives a good fit to the data points.

Part B
What is the initial speed v0 of the particle when it is very far from the sphere?
Express your answer to three significant figures and include the appropriate units.

Part C
What is Q?
Express your answer to three significant figures and include the appropriate units.

Part D
How close does the particle get to the sphere? Assume that this distance is much larger than the radii of the particle and sphere, so continue to treat them as point particles and to assume that the sphere remains at rest.
Express your answer to three significant figures and include the appropriate units.

 

.

 

 

 

v2 (m2/s2)
350
300
250
200
50
100
50
(m)
0
2 4 6 8 10 12 14 16 18
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v2 (m2/s2) 350 300 250 200 50 100 50 (m) 0 2 4 6 8 10 12 14 16 18

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Expert Answer

Step 1

Given information:

Mass of the particle (m) = 4 × 10-4 kg

Charge on the particle (q) = 5 × 10-8 C

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Equation that describes the motion V 2 (15.75 m2 s2

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Step 2

The force exerted by the stationary charge (Q) on the particle is of charge (q):

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kQq F = ma = х2 hence the acceleration of the particle is given by: kQq а x2m Let "u" be the initial velocity of the particle, hence from 3rd equation of motion we can write: V2 u2|(2x)(q V2 —D и? + Equation (1 хт

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Step 3

Part B:

Compare equation 1 with given equation of motion:

We can se that u2 = 400 m2/s2 → u = 20 m/s

Hence init...

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SciencePhysics

Electric Charges and Fields

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