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Chapter 17: Problem 39

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 mass of the drop is \(1.0 \times 10^{-13} \mathrm{kg}\) and it remains stationary when the potential difference between the plates is $9.76 \mathrm{kV},$ what is the magnitude of the charge on the drop? (Ignore the small buoyant force on the drop.)

Short Answer

Expert verified
Answer: The magnitude of the charge on the oil drop is approximately 1.61 x 10^-17 C.

Step by step solution

01

Determine the gravitational force on the oil drop

Using the given mass of the oil drop (\(m = 1.0 \times 10^{-13} \mathrm{kg}\)), we can find the gravitational force acting on it. The gravitational force is given by: \(F_g = m \cdot g\) where \(g\) is the acceleration due to gravity and is approximately equal to \(9.81 \mathrm{m/s^2}\). Therefore, the gravitational force acting on the oil drop is: \(F_g = (1.0 \times 10^{-13} \mathrm{kg})(9.81 \mathrm{m/s^2}) = 9.81 \times 10^{-13} \mathrm{N}\).
02

Determine the electric field between the plates

We are given that the potential difference between the plates is \(9.76 \mathrm{kV}\), and the distance between the plates is \(16 \mathrm{cm}\). We can convert the distance to meters (\(0.16 \mathrm{m}\)) and the potential difference to volts (\(9760 \mathrm{V}\)). The electric field (\(E\)) between the plates can be calculated using the formula: \(E = \dfrac{V}{d}\) where \(V\) is the potential difference, and \(d\) is the distance between the plates. So, we find: \(E = \dfrac{9760 \mathrm{V}}{0.16 \mathrm{m}} = 61000 \mathrm{V/m}\)
03

Calculate the electric force on the oil drop

The electric force acting on the oil drop can be calculated using the formula: \(F_e = E \cdot q\) where \(F_e\) is the electric force, \(E\) is the electric field, and \(q\) is the charge on the oil drop. Since the oil drop is stationary, the electric force must be equal to the gravitational force: \(F_e = F_g\) Substituting the equation for the electric force, we get: \(E \cdot q = F_g\)
04

Solve for the magnitude of the charge on the oil drop

Now that we have found \(E\) and \(F_g\), we can solve for \(q\): \(q = \dfrac{F_g}{E} = \dfrac{9.81 \times 10^{-13} \mathrm{N}}{61000 \mathrm{V/m}} = 1.61 \times 10^{-17} \mathrm{C}\) So, the magnitude of the charge on the oil drop is approximately \(1.61 \times 10^{-17} \mathrm{C}\).

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Most popular questions from this chapter

The capacitor of Problem 79 is initially charged to a \(150-\mathrm{V}\) potential difference. The plates are then physically separated by another \(0.750 \mathrm{mm}\) in such a way that none of the charge can leak off the plates. Find (a) the new capacitance and (b) the new energy stored in the capacitor. Explain the result using conservation of energy.
Two parallel plates are \(4.0 \mathrm{cm}\) apart. The bottom plate is charged positively and the top plate is charged negatively, producing a uniform electric field of \(5.0 \times 10^{4} \mathrm{N} / \mathrm{C}\) in the region between the plates. What is the time required for an electron, which starts at rest at the upper plate, to reach the lower plate? (Assume a vacuum exists between the plates.)
A spherical conductor with a radius of \(75.0 \mathrm{cm}\) has an electric field of magnitude \(8.40 \times 10^{5} \mathrm{V} / \mathrm{m}\) just outside its surface. What is the electric potential just outside the surface, assuming the potential is zero far away from the conductor?
A parallel plate capacitor has a capacitance of \(1.20 \mathrm{nF}\) and is connected to a \(12-\mathrm{V}\) battery. (a) What is the magnitude of the charge on each plate? (b) If the plate separation is doubled while the plates remain connected to the battery, what happens to the charge on each plate and the electric field between the plates?
In a region where there is an electric field, the electric forces do $+8.0 \times 10^{-19} \mathrm{J}\( of work on an electron as it moves from point \)X$ to point \(Y\). (a) Which point, \(X\) or \(Y\), is at a higher potential? (b) What is the potential difference, \(V_{Y}-V_{X},\) between point \(Y\) and point \(X ?\)
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