Copyright

Three similar metal spheres of the same radius and mass on insulated stands are initially charged...

Question:

Three similar metal spheres of the same radius and mass on insulated stands are initially charged as follows: sphere A has {eq}+6.0 \ nC {/eq} charge, sphere B has {eq}-2.0 \ nC {/eq} of charge, and sphere C is neutral. In this order, the spheres are touched for long enough for the charges to come to equilibrium and then separated again. Sphere A is in contact with B and then separates. Sphere A then touches C (and then separates). Sphere C then touches B (and then separates).

a. Find the net electric field at {eq}\langle 0.0, -1.0\rangle \ cm. {/eq}

b. If a {eq}-3 \ \mu C {/eq} is placed at {eq}\langle 0.0, -1.0\rangle \ cm {/eq}, what would the electrostatic force on this charge be?

Electrostatic balance


Since electrons in electric conductors can move freely within the material, then by contacting 2 or more conductors they will share their electrons. When bodies have some kind of excess charge (positive or negative), the exchange of electrons will lead to the electrical potentials of the bodies in contact being equal. This is what is known as electrostatic balance.


Answer and Explanation:


Because in this case the spheres are identical, after being separated they will have the same charge:

{eq}\textbf{Initial charge}\\ q_{A0}=6.0\,\mathrm{nC}\\ q_{B0}=-2.0\,\mathrm{nC}\\ q_{B0}=0\,\mathrm{C}\\ \textbf{Contact A-B}\\ \text{Before }\rightarrow \,q_{A0}+q_{B0}=4.0\,\mathrm{nC}\\ \displaystyle \text{After }\rightarrow \,q_{A1}=q_{B1}=\frac{q_{A0}+q_{B0}}{2}=2.0\,\mathrm{nC}\\ \textbf{Contact A-C}\\ \text{Before }\rightarrow \,q_{A1}+q_{C0}=2.0\,\mathrm{nC}\\ \displaystyle \text{After }\rightarrow \,q_{A2}=q_{C1}=\frac{q_{A1}+q_{C0}}{2}=1.0\,\mathrm{nC}\\ \textbf{Contact B-C}\\ \text{Before }\rightarrow \,q_{B1}+q_{C1}=3.0\,\mathrm{nC}\\ \displaystyle \text{After }\rightarrow \,q_{B2}=q_{C2}=\frac{q_{B1}+q_{C1}}{2}=1.5\,\mathrm{nC}\\ \textbf{Final charge}\\ q_{A2}=1.0\,\mathrm{nC}\\ q_{B2}=1.5\,\mathrm{nC}\\ q_{C2}=1.5\,\mathrm{C}\\ {/eq}

In this case we do not know the positions of spheres A, B and C, therefore it is impossible to find exactly the expression of the electric field and the electrostatic force exerted by these charges at any point of space. This is conditioned by the fact that the direction of the electric field at any point depends on the relative positions of the spheres.


a. Electric field (generic case)

{eq}\vec{E}(x,y)=\vec{E}_A(x,y)+\vec{E}_B(x,y)+\vec{E}_C(x,y)\\ \vec{E}(x,y)=(E_{Ax}+E_{Bx}+E_{Cx})\hat{i}+(E_{Ay}+E_{By}+E_{Cy})\hat{j}\\ \displaystyle E_{ix}=\frac{kq_i}{(x_i-x)^2+(y_i-y)^2}\cos\theta_i\\ \displaystyle E_{iy}=\frac{kq_i}{(x_i-x)^2+(y_i-y)^2}\sin\theta_i\\ {/eq}

where {eq}q_i {/eq} is the electric charge of the spheres A, B and C that are located at coordinates {eq}(x_i,y_i) {/eq}, while {eq}\theta_i {/eq} is the angle that the electric field makes counterclockwise with respect to the positive direction of the x axis.


b. Electrostatic force (general case) acting on any charge {eq}q {/eq}:

{eq}\vec{F}(x,y)=q\vec{E}(x,y)\\ {/eq}

where {eq}\vec{E}(x,y) {/eq} is the electric field found in the previous section.


Learn more about this topic:

Loading...
Electric Fields: Definition & Examples

from General Studies Science: Help & Review

Chapter 3 / Lesson 5
66K

Related to this Question

Explore our homework questions and answers library