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Asked Nov 6, 2019
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Ch 18, Problem 44

A spring with an unstrained length of 0.071 m and a spring constant of 2.4 N/m hangs vertically downward from the ceiling. A uniform electric field directed upward fills the region containing the spring. A sphere with a mass of 5.0 × 10-3 kg and a net charge of +6.6 μC is attached to the lower end of the spring. The spring is released slowly, until it reaches equilibrium. The equilibrium length of the spring is 0.058 m. What is the magnitude of the external electric field?

L
Lo
Unstrained spring
Spring with
charged sphere
and electric field
HММИК
ИтЕ
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L Lo Unstrained spring Spring with charged sphere and electric field HММИК ИтЕ

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Expert Answer

Step 1

The forces acting on the system are the electric force, spring force and the weight of the sphere. The weight of the sphere acts downward whereas the spring force and the electric force act upward. Taking the upward direction to be positive, the force equation at equilibrium is given by

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Here, F is the electric force, F is the spring force and W is the weight of the sphere. The electric force is given by Here, q is the charge on the sphere and E is the electric field. The equation for the spring force is given by Here, k is the spring constant and Ay is the change in the length of the spring The equation for the weight of the sphere is given by = mg Here, mis the mass of the sphere and g is the acceleration due to gravity Put the above three equations in equation (1) qE+(-kAy)mg qE-kAy mg qE mg kAy kAy (2) mg E =

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Step 2

The change in the length of t...

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Ay 0.071 m -0.058 m = 0.013 m Substitute the numerical values in equation (2) (5.0x10-3 kg) (9.80 m/s)+ (2.4 N/m)(0.013 m) E = 6.6x10 C 1.22 x 104 N/C

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