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Chapter 16: Problem 90

A very small charged block with a mass of 2.35 g is placed on an insulated, frictionless plane inclined at an angle of \(17.0^{\circ}\) with respect to the horizontal. The block does not slide down the plane because of a \(465-\mathrm{N} / \mathrm{C}\) uniform electric field that points parallel to the surface downward along the plane. What is the sign and magnitude of the charge on the block?

Short Answer

Expert verified
Answer: The sign and magnitude of the charge on the block are approximately -7.73 × 10⁻⁶ C.

Step by step solution

01

Draw a free body diagram

Draw a free body diagram of the block on the inclined plane. Include the gravitational force \(mg\) acting downward, the normal force \(N\) acting perpendicular to the inclined plane, and the electric force \(F_e\) acting parallel to the inclined plane.
02

Decompose the gravitational force into components

Decompose the gravitational force \(F_g\) into two components, \(F_{gx}\) and \(F_{gy}\), where \(F_{gx}\) is parallel to the inclined plane, and \(F_{gy}\) is perpendicular to the inclined plane. Remember that gravitational force acting on the block is: $$F_g = mg$$ From the angle of inclination \(17.0^{\circ}\) , components can be found as: $$F_{gx} = F_g \sin{\theta} = mg \sin{17.0^{\circ}}$$ $$F_{gy} = F_g \cos{\theta} = mg \cos{17.0^{\circ}}$$
03

Write Newton's Second Law for the block

Apply Newton's Second Law of Motion equation, \(F = ma\) on the block along the inclined direction of the plane. $$F_{gx} = F_e$$ Since the electric field \(E\) is given and the electric force \(F_e\) is calculated using the formula: $$F_e = qE$$ We can rewrite Newton's Second Law for this case as: $$mg \sin{17.0^{\circ}} = qE$$
04

Solve for the charge \(q\)

Now solve for the charge \(q\) by plugging in the known values for the mass \(m\), gravitational constant \(g\), angle \(\theta\), and electric field \(E\). $$q = \frac{mg \sin{17.0^{\circ}}}{E}$$ Plug in the values: \(m=0.00235\,\text{kg}\), \(g=9.81\,\text{m/s}^2\), and \(E=465\,\text{N/C}\). $$q = \frac{(0.00235 \,\text{kg})(9.81\, \text{m/s}^2)\sin{17.0^{\circ}}}{ 465\, \text{N/C}}$$ $$q \approx 7.73 \times 10^{-6}\,\text{C}$$
05

Determine the sign of the charge

Since the electric field and electric force both point parallel to the surface downward along the inclined plane, the force opposes the block's motion. Therefore, the charge on the block must have the same sign as the electric field (if the charge and field had opposite signs, the electric force would act in the opposite direction and the block would slide down the plane). In this case, the electric field points downward, which is considered a negative direction. Thus, the charge of the block is also negative.
06

Final Answer

The sign and magnitude of the charge on the block are: $$q \approx -7.73 \times 10^{-6}\,\text{C}$$

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Most popular questions from this chapter

An object with a charge of \(0.890 \mu \mathrm{C}\) is placed at the center of a cube. What is the electric flux through one surface of the cube?
Point charges are arranged on the vertices of a square with sides of $2.50 \mathrm{cm} .$ Starting at the upper left corner and going clockwise, we have charge A with a charge of \(0.200 \mu \mathrm{C}, \mathrm{B}\) with a charge of \(-0.150 \mu \mathrm{C}, \mathrm{C}\) with a charge of \(0.300 \mu \mathrm{C},\) and \(\mathrm{D}\) with a mass of \(2.00 \mathrm{g},\) but with an unknown charge. Charges \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\) are fixed in place, and \(\mathrm{D}\) is free to move. Particle D's instantaneous acceleration at point \(D\) is \(248 \mathrm{m} / \mathrm{s}^{2}\) in a direction \(30.8^{\circ}\) below the negative \(x\) -axis. What is the charge on D?
Positive point charges \(q\) and \(2 q\) are located at \(x=0\) and \(x=3 d,\) respectively. What is the electric field at \(x=2 d\) (point \(S\) )?
A thin, flat sheet of charge has a uniform surface charge density \(\sigma(\sigma / 2\) on each side). (a) Sketch the field lines due to the sheet. (b) Sketch the field lines for an infinitely large sheet with the same charge density. (c) For the infinite sheet, how does the field strength depend on the distance from the sheet? [Hint: Refer to your field line sketch.J (d) For points close to the finite sheet and far from its edges, can the sheet be approximated by an infinitely large sheet? [Hint: Again, refer to the field line sketches.] (e) Use Gauss's law to show that the magnitude of the electric field near a sheet of uniform charge density \(\sigma\) is $E=\sigma /\left(2 \epsilon_{0}\right)$
A charge of \(63.0 \mathrm{nC}\) is located at a distance of \(3.40 \mathrm{cm}\) from a charge of \(-47.0 \mathrm{nC} .\) What are the \(x\) - and \(y\) -components of the electric field at a point \(P\) that is directly above the 63.0 -nC charge at a distance of \(1.40 \mathrm{cm} ?\) Point \(P\) and the two charges are on the vertices of a right triangle.
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