Question

Your lab assignment for the week is to measure the amount of charge on the 6.0-cm-diameter metal sphere of a Van de Graaff generator. To do so, you’re going to use a spring with a spring constant of 0.65 N/m to launch a small, 1.5 g bead horizontally toward the sphere. You can reliably charge the bead to 2.5 nC, and your plan is to use a video camera to measure the bead’s closest approach to the edge of the sphere as you change the compression of the spring. Your data is as follows:

Use an appropriate graph of the data to determine the sphere’s charge in nC. You can assume that the bead’s motion is entirely horizontal, that the spring is so far away that the bead has no interaction with the sphere as it’s launched, and that the approaching bead does not alter the charge distribution on the sphere.

Short answer

The charge on sphere is 310nC.

Step-by-step solution

Step 1. Concept use

Electric potential energy is a potential energy (measured in joules) resulting from conservative Coulomb forces and associated with the configuration of a specific set of point charges within a defined system.

Step 2. Simplify

When we compress the spring the distance A, the potential energy stored in the spring is equal to

$\mathrm{PE}\text{elastic}=\frac{1}{2}k{A}^2\text{,}$

Where$k=0.65\mathrm{N}/\mathrm{m}$.

In the final moment , when the bead stops, its total energy is equal to its electric potential energy:

${\mathrm{PE}}_{\text{electric}}=\frac{1}{4\pi {\epsilon }_0}\frac{qQ}{d+R},$

Where q is the charge of the bead, Q is the unknown charge of the sphere is $d+R$and d is the distance of closest approach (the total distance from the center of the sphere is where R=3 cm

From energy conservation we find

$\frac{1}{2}k{A}^2=\frac{1}{4\pi {\epsilon }_0}\frac{qQ}{d+R}$

We can write this as

$A^2=\frac{qQ}{2\pi {\epsilon }_{0}k}·\frac{1}{d+R}$.

Step 3.

In the previous equation we can set $y=A^2$and x=1 /(d+R) to get the linear relation

$y=ax$

Where

$a=\frac{qQ}{2\pi {\epsilon }_{0}k}.$

We can graph this linear dependence y(x) and determine a from that graph. The required data is given in the table bellow.

Step 4. Finding the charge

The fitted graph (black line) of the data given in the table (red dots) is shown bellow.

To find the coefficient a we use two nonexperimental points on the graph - on of them between the first to experimental points and another one between the last two. Their coordinates are

$P_1:x_1=0.141{\mathrm{cm}}^{-1},y_1=3{\mathrm{cm}}^2,$

and

$P_2:x_2=0.250{\mathrm{cm}}^{-1},y_2=5.33{\mathrm{cm}}^2$

Now the slope a is found as

$a=\frac{y_2-y_1}{x_2-x_1}=21.4{\mathrm{cm}}^3$

Now solving for Q we have that,

$Q=\frac{2\pi {\epsilon }_{0}ka}{q}=310\mathrm{nC}$.