Chapter 17: Problem 38
A positively charged oil drop is injected into a region of uniform electric field between two oppositely charged, horizontally oriented plates spaced $16 \mathrm{cm}$ apart. If the electric force on the drop is found to be $9.6 \times 10^{-16} \mathrm{N}\( and the potential difference between the plates is \)480 \mathrm{V}$ what is the magnitude of the charge on the drop in terms of the elementary charge \(e\) ? Ignore the small buoyant force on the drop.
Short Answer
Expert verified
#tag_title#Step 2: Calculate the electric field#tag_content#
Now, we can plug the values into the formula and calculate the electric field:
$$
E = \frac{480 \mathrm{V}}{0.16 \mathrm{m}} = 3000 \frac{\mathrm{V}}{\mathrm{m}}
$$
The electric field between the plates is 3000 V/m.
#tag_title#Step 3: Calculate the charge on the oil drop#tag_content#
We can use the formula for electric force:
$$
F = Eq
$$
Where F is the electric force acting on the charge, E is the electric field, and q is the charge.
We are given F = 2.5 x 10^(-15) N and E = 3000 V/m. We need to find the charge, q, so we can rearrange the formula to solve for q:
$$
q = \frac{F}{E}
$$
Now, plug in the values:
$$
q = \frac{2.5 \times 10^{-15} \mathrm{N}}{3000 \frac{\mathrm{V}}{\mathrm{m}}}
$$
#tag_title#Step 4: Calculate the charge in terms of elementary charge#tag_content#
Now, we can calculate the value of q:
$$
q = \frac{2.5 \times 10^{-15} \mathrm{N}}{3000 \frac{\mathrm{V}}{\mathrm{m}}} = 8.33 \times 10^{-19} \mathrm{C}
$$
To express the charge in terms of elementary charge (e = 1.6 x 10^(-19) C), we can divide the charge, q, by the elementary charge:
$$
\frac{8.33 \times 10^{-19} \mathrm{C}}{1.6 \times 10^{-19} \mathrm{C}} = 5.2
$$
This means the charge on the oil drop is approximately 5.2 times the elementary charge.
Step by step solution
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