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Chapter 22: Problem 9

Three \(-9-\mathrm{mC}\) point charges are located at (0,0) \((3 \mathrm{~m}, 3 \mathrm{~m})\), and \((3 \mathrm{~m},-3 \mathrm{~m})\). What is the magnitude of the electric field at \((3 \mathrm{~m}, 0) ?\) a) \(0.9 \cdot 10^{7} \mathrm{~N} / \mathrm{C}\) b) \(1.2 \cdot 10^{7} \mathrm{~N} / \mathrm{C}\) c) \(1.8 \cdot 10^{7} \mathrm{~N} / \mathrm{C}\) d) \(2.4 \cdot 10^{7} \mathrm{~N} / \mathrm{C}\) e) \(3.6 \cdot 10^{7} \mathrm{~N} / \mathrm{C}\) f) \(5.4 \cdot 10^{7} \mathrm{~N} / \mathrm{C}\) g) \(10.8 \cdot 10^{7} \mathrm{~N} / \mathrm{C}\)

Short Answer

Expert verified
a) \(0.9 \cdot 10^{7} \mathrm{~N} / \mathrm{C}\) b) \(2.7 \cdot 10^{7} \mathrm{~N} / \mathrm{C}\) c) \(9 \cdot 10^{6} \mathrm{~N} / \mathrm{C}\) d) \(0\mathrm{~N} / \mathrm{C}\) Answer: a) \(0.9 \cdot 10^{7} \mathrm{~N} / \mathrm{C}\)

Step by step solution

01

Recall the formula for electric field due to a point charge

The electric field E due to a point charge Q at a distance r from the charge is given by the formula: \[ E = \frac{kQ}{r^2} \] Where k is the electrostatic constant, approximately \(8.99 \cdot 10^{9} \frac{\mathrm{N} \cdot \mathrm{m^2}}{\mathrm{C^2}}\).
02

Calculate the distance from each charge to the point of interest

First, let's find the distance from each of the charges to the point (3 m, 0), by using the distance formula in two-dimensional space: \[r = \sqrt{(x-x_0)^2 + (y-y_0)^2}\] For the charge at (0, 0): \[r_1 = \sqrt{(3-0)^2 + (0-0)^2} = 3\text{ m}\] For the charge at (3 m, 3 m): \[r_2 = \sqrt{(3-3)^2 + (0-3)^2} = 3\text{ m}\] For the charge at (3 m, -3 m): \[r_3 = \sqrt{(3-3)^2 + (0+3)^2} = 3\text{ m}\]
03

Calculate the electric field for each point charge

Now, we will use the electric field formula and the calculated distances to find the electric field due to each charge at the point (3 m, 0). For the charge at (0, 0): \[E_1 = \frac{k(-9 \cdot 10^{-9} \text{C})}{(3\text{ m})^2} = -9\cdot10^6\text{ N/C along the x-axis}\] For the charge at (3 m, 3 m): \[E_2 = \frac{k(-9 \cdot 10^{-9} \text{C})}{(3\text{ m})^2} = -9\cdot10^6\text{ N/C along the -y-axis (downward)}\] For the charge at (3 m, -3 m): \[E_3 = \frac{k(-9 \cdot 10^{-9} \text{C})}{(3\text{ m})^2} = -9\cdot10^6\text{ N/C along the y-axis (upward)}\]
04

Find the vector sum of the electric fields

Next, we need to find the vector sum of these electric fields. Since \(E_2\) and \(E_3\) are along -y and y axes respectively with equal magnitudes, they will cancel each other out. Thus, the total electric field is along the x-axis and has a magnitude of \(-9 \cdot 10^6\text{ N/C}\) (which is the strength of \(E_1\))
05

Determine the magnitude of the total electric field

Since the electric field is entirely along the x-axis, the magnitude will be the same as its x-component (ignoring the negative sign): \[E_{total} = |-9 \cdot 10^6\text{ N/C}| = 9 \cdot 10^6\text{ N/C}\] After comparing this result with the given choices, the correct answer is: a) \(0.9 \cdot 10^{7} \mathrm{~N} / \mathrm{C}\).

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Most popular questions from this chapter

How is it possible that the flux through a closed surface does not depend on where inside the surface the charge is located (that is, the charge can be moved around inside the surface with no effect whatsoever on the flux)? If the charge is moved from just inside to just outside the surface, the flux changes discontinuously to zero, according to Gauss's Law. Does this really happen? Explain.

Repeat Example 22.3 , assuming that the charge distribution is \(-\lambda\) for \(-a

A water molecule, which is electrically neutral but has a dipole moment of magnitude \(p=6.20 \cdot 10^{-30} \mathrm{C} \mathrm{m},\) is \(1.00 \mathrm{~cm}\) away from a point charge \(q=+1.00 \mu \mathrm{C} .\) The dipole will align with the electric field due to the charge. It will also experience a net force, since the field is not uniform. a) Calculate the magnitude of the net force. (Hint: You do not need to know the precise size of the molecule, only that it is much smaller than \(1 \mathrm{~cm} .)\) b) Is the molecule attracted to or repelled by the point charge? Explain.

A long conducting wire with charge distribution \(\lambda\) and radius \(r\) produces an electric field of \(2.73 \mathrm{~N} / \mathrm{C}\) just outside the surface of the wire. What is the magnitude of the electric field just outside the surface of another wire with charge distribution \(0.81 \lambda\) and radius \(6.5 r ?\)

Electric dipole moments of molecules are often measured in debyes \((\mathrm{D}),\) where \(1 \mathrm{D}=3.34 \cdot 10^{-30} \mathrm{C} \mathrm{m} .\) For instance, the dipole moment of hydrogen chloride gas molecules is \(1.05 \mathrm{D}\). Calculate the maximum torque such a molecule can experience in the presence of an electric field of magnitude \(160.0 \mathrm{~N} / \mathrm{C}\).
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