A 2.0 mm diameter copper ball is charged to +42 nC. What fraction of its electrons have been removed? The density of copper is 8,920 kg/m3
around 14% (make it a fraction)
Yeah
To find the fraction of electrons that have been removed from the copper ball, we need to determine the total number of electrons in the ball and compare it to the number of electrons that remain after the removal.
To find the total number of electrons in the ball, we can use the formula:
Total number of electrons = (Total charge) / (Charge of one electron)
The charge of one electron is given by the elementary charge, which is approximately -1.6 x 10^-19 C.
First, we need to convert the diameter of the ball from millimeters to meters:
Diameter = 2.0 mm = 0.002 m
Next, we can calculate the volume of the ball using its diameter:
Volume = (4/3) * π * (radius)^3
Since the diameter is given, we can find the radius by dividing it by 2:
Radius = Diameter / 2
Now we can substitute the radius into the volume formula to find the volume of the ball.
Next, we need to determine the mass of the copper ball. We can use the density of copper and the volume of the ball to calculate this.
Mass = Volume * Density
Finally, we can calculate the total charge of the ball using the charge and the charge per electron formula mentioned earlier.
Total charge = Number of electrons x Charge of one electron
Now we can compare the total number of electrons with the charge removed from the ball (+42 nC) to find the fraction of electrons removed.
Fraction of electrons removed = (Charge removed) / (Total charge)
Let's go through the steps to find the fraction of electrons removed from the copper ball:
Step 1: Calculate the volume of the copper ball.
Step 2: Determine the mass of the copper ball.
Step 3: Calculate the total charge of the ball.
Step 4: Use the total charge and the charge removed to find the fraction of electrons removed.
Please provide the density of copper so we can proceed with the calculations.