1) A small glass bead has been charged to +17nC. A metal ball bearing 1.0cm above the bead feels...

Question:

1) A small glass bead has been charged to +17nC. A metal ball bearing 1.0cm above the bead feels a 0.030N downward electric force. What is the charge on the ball bearing?

2) Objects A and B are both positively charged. Both have a mass of 92g, but A has twice the charge of B. When A and B are placed 10cm apart, B experiences an electric force of 0.45N.

a) How large is the force on A?

b) What are the charges qA and qB?

c) If the objects are released, what is the initial acceleration of A?

Electrostatic Force Between Charged Objects

Electrostatic force between any two objects can be calculated using Coulomb's law by considering the objects as point charges and distance of separation as the distance between the centers of the objects. In this calculation we assume that the whole charge on the object is centered at the center of the object. Electrostatic force can be attractive or repulsive but its magnitude {eq}F = \dfrac { k q_1 q_2 } { d^2 } {/eq}. Here {eq}k, \ \ q_1, \ \ q_2, \ \ d {/eq} are the Coulomb constant, charge on object one, charge on object two, distance of separation between the objects respectively.

Answer and Explanation:

Q1

  • Charge on the glass bead {eq}q_1 = + 17.0 \times 10^{-9 } \ \ C {/eq}
  • Distance to the ball baring {eq}d = 1.00 \times 10^{-2 } \ \ m {/eq}
  • Electrostatic force of attraction between the glass bead and metal ball baring {eq}F = 0.030 \ \ N {/eq}
  • Coulomb constant {eq}k = 8.99 \times 10^9 \ \ N m^2/C^2 {/eq}

Let {eq}q_2 {/eq} be the charge on the ball baring

Then the magnitude of the charge on the baring {eq}q_2 = \dfrac { F d^2 } { k q_1 } \\ = \dfrac { 0.030 \times ( 1.0 \times 10^{-2 } )^2} { 8.99 \times 10^9 \times 17.0 \times 10^{-9} } \\ = 19.63 \times 10^{-9} \ \ C {/eq}

This charge must be negative as the force is of attractive in nature.

Q2

  • Objects A and B are positively charged.
  • Mass of each object {eq}m = 92.0 \times 10^{-3 } \ \ kg {/eq}
  • Distance of separation between the charges d = 0.10 m
  • Charge on object B is q , then charge on object A is 2q.
  • Electric force experienced by object B , F = 0.45 N

Part a)

According to Coulomb's law both the charges involved in the Coulomb's law experience the same force.

So force experienced by object A is also F = 0.45 N

Part b)

Electrostatic force between the objects A and B , {eq}F = k \times \dfrac { q 2q } { d^2 } {/eq}

So charge on object B is {eq}q_B = \sqrt { \dfrac { F d^2 } { 2 k } } \\ = \sqrt { \dfrac { 0.45 \times 0.10^2 } { 2 \times 8.99 \times 10^9 } } \\ = 5.003 \times 10^{-7 } \ \ C {/eq}

So charge on object A is {eq}q_A = 2 \times q_B \\ = 2 \times 5.003 \times 10^{-7 } \\ = 1.001 \times 10^{-6 } \ \ C {/eq}

Part c)

Initial acceleration of object A , {eq}a = \dfrac { F } { m } \\ = \dfrac { 0.45 } { 92.0 \times 10^{-3 } } \\ = 4.891 \ \ m/s^2 {/eq}


Learn more about this topic:

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Coulomb's Law: Variables Affecting the Force Between Two Charged Particles

from CLEP Natural Sciences: Study Guide & Test Prep

Chapter 6 / Lesson 3
80K

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