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Chapter 9: Problem 24

Two negative charges with magnitude \(q\) and \(2 q\) sit at points \((1,0)\) and \((0,1)\) on the \(x\) - and \(y\)-axis, respectively. Which figure best represents the total electric field at the origin? (A) \(\mathrm{A}\) (B) \(\mathrm{B}\) (C) \(\mathrm{C}\) (D) \(\mathrm{D}\) (E) \(\mathrm{E}\)

Short Answer

Expert verified
Question: Describe the direction and magnitude of the total electric field at the origin, considering the presence of a negative charge \(q\) at (1,0) and a negative charge \(2q\) at (0,1). Answer: The total electric field at the origin has a direction between the negative x and y-axes, with a stronger field towards the negative y-axis due to the presence of \(2q\). The magnitude along the x-axis is \(kq\) and along the y-axis is \(2kq\).

Step by step solution

01

Find the electric field due to charge \(q\) at the origin.

First, let's find the electric field at the origin due to the charge \(q\) at point (1,0). The direction of the electric field is always away from a positive charge and towards a negative charge. Since the charge at point (1,0) is negative, the electric field at the origin due to this charge will be directed towards it, i.e., along the negative x-axis. The magnitude of the electric field due to this charge, \(E_q\), can be calculated using Coulomb's law: \(E_q = \frac{kq}{r^2}\) Here, \(k\) is Coulomb's constant, \(q\) is the charge magnitude, and \(r\) is the distance between the charge and the origin. In this case, \(r = 1\), so the electric field magnitude due to this charge is \(E_q = kq.\)
02

Find the electric field due to charge \(2q\) at the origin.

Now let's find the electric field at the origin due to the charge \(2q\) at point (0,1). Since this charge is also negative, the electric field at the origin due to this charge will be directed towards it, i.e., along the negative y-axis. The magnitude of the electric field due to this charge, \(E_{2q}\), can be calculated using Coulomb's law: \(E_{2q} = \frac{k(2q)}{r^2}\) Again, the distance between the charge and the origin is \(r = 1\), so the electric field magnitude due to this charge is \(E_{2q} = 2kq.\)
03

Find the vector sum of the individual electric fields.

We can now find the total electric field at the origin by calculating the vector sum of the electric fields due to each charge. \(E_{total} = \vec{E_q} + \vec{E_{2q}}\) Since the electric field due to the charge at (1,0) is directed along the negative x-axis and the electric field due to the charge at (0,1) is directed along the negative y-axis, we can write the electric field vector components for these charges: \(\vec{E_q} = (-kq, 0)\) \(\vec{E_{2q}} = (0, -2kq)\) \(E_{total} = \vec{E_q} + \vec{E_{2q}} = (-kq, 0) + (0, -2kq) = (-kq, -2kq)\)
04

Describe the total electric field at the origin.

The total electric field at the origin, \(E_{total}\), is given by the vector \((-kq, -2kq)\). This means that the electric field has a magnitude along the x-axis of \(kq\) and a magnitude twice as large along the y-axis (\(2kq\)). In other words, the electric field is pointing in the direction between the negative x and y-axes, with a stronger electric field towards the negative y-axis. The best representation of the total electric field will have a direction between the negative x and y-axes, with a stronger electric field towards the negative y-axis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coulomb's Law
Coulomb's Law helps us understand how electric fields are generated by point charges. It's a fundamental law of electromagnetism that uses the formula \[ E = \frac{kq}{r^2} \]where:
  • \(E\) is the electric field,
  • \(k\) is Coulomb's constant (approximately \(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),
  • \(q\) is the charge magnitude,
  • \(r\) is the distance from the charge.
The direction of the electric field depends on the sign of the charge:
  • Positive charges generate pointing away.
  • Negative charges create fields directed toward them.
In our exercise, there are two negative charges. For the first charge located at point (1,0), the field at the origin points towards the negative x-axis, calculated as \(E_q = kq\) due to a distance \(r\) of 1 unit. For the second charge at (0,1), the magnitude is \(E_{2q} = 2kq\) along the negative y-axis. Coulomb's Law offers clarity on getting these electric field magnitudes.
Vector Addition
When we need to find a total effect from multiple sources of electric fields, vector addition becomes vital. Electric fields are vector quantities, meaning they have both magnitude and direction. In this problem, we combine two vectors:
  • The field from charge \(q\): \(\vec{E_q} = (-kq, 0)\)
  • The field from charge \(2q\): \(\vec{E_{2q}} = (0, -2kq)\)
We perform vector addition by summing the components in each direction. Here, \[ E_{total} = \begin{bmatrix} -kq \ 0 \end{bmatrix} + \begin{bmatrix} 0 \ -2kq \end{bmatrix} = \begin{bmatrix} -kq \ -2kq \end{bmatrix} \].This result shows that the total electric field has both an x and a y component, pointing in a direction diagonally between the negative x and negative y-axes. It's crucial to correctly add vectors to determine the proper direction and magnitude of resultant fields.
Charge Interaction
Understanding charge interactions help explain the behavior of the electric field. Charges exert forces on each other and generate electric fields:
  • Like charges (both positive or both negative) repel each other.
  • Unlike charges attract.
In this exercise, both charges at (1,0) and (0,1) are negative, which impact how the electric fields they produce interact at a point like the origin. These interactions determine the strength and direction of the resultant electric field:
  • For two negative charges, both point their electric fields towards themselves from any external point, exemplifying charge repulsion.
  • The interaction at the origin results in a stronger influence along the y-axis since one charge has doubled magnitude.
Thus, understanding charge interactions is key to predicting the resultant forces and directions in electric field scenarios.

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Most popular questions from this chapter

Proton \(A\) is moving with speed \(10^6 \mathrm{~m} / \mathrm{s}\) in a magnetic field of \(0.01 \mathrm{~T}\). Proton \(B\) is moving in the same magnetic field with speed \(2 \times 10^6 \mathrm{~m} / \mathrm{s}\). One can conclude that (A) Both protons have an orbital radius of about 1 meter. (B) Both protons have the same orbital period of about \(6.5 \times 10^{-6} \mathrm{~s}\). (C) proton \(B\) has twice the orbital period of proton \(A\). (D) proton \(B\) has twice the orbital frequency of proton \(A\). (E) proton \(B\) has half the orbital radius of proton \(B\).

Two identical point charges \(q_1\) and \(q_2\) are at a distance \(r\) apart. If the size of \(q_1\) is doubled and the distance between them tripled, the strength of the electrical force between them (A) goes up by a factor of 3 . (B) goes down by a factor of 3 . (C) goes down by a factor of 9 . (D) goes down by a factor of \(2 / 3\). (E) goes down by a factor of \(2 / 9\)

Two long, parallel wires separated by a distance \(r\) carry equal currents \(I\) in opposite directions, as shown. The direction of the field caused by the top wire at the position of the bottom wire and the direction of the force exerted by the top wire on the bottom wire are (A) \(B\) into the page; \(F\) down (B) \(B\) up; \(F\) into the page (C) \(B\) into the page; \(F\) up (D) \(B\) out of the page; \(F\) down (E) \(B\) down; \(F\) out of the page

Two positive charges of magnitude \(q\) and \(2 q\) are fixed in place along the \(x\)-axis. Is there any place along the \(x\)-axis where the total field could be zero? (A) Yes, somewhere to the left of the charge \(q\) (B) Yes, somewhere to the right of the charge \(2 q\) (C) Yes, between the two charges but closer to \(q\) (D) Yes, between the two charges but closer to \(2 q\) (E) No, the field can never be zero

Metal sphere \(A\) has a radius of \(5 \mathrm{~cm}\) and metal sphere \(B\) has a radius of \(10 \mathrm{~cm}\). Sphere \(A\) carries a charge of \(9 \mathrm{nC}\) and sphere \(B\) carries a charge of \(18 \mathrm{nC}\). If the surfaces of \(A\) and \(B\) are \(185 \mathrm{~cm}\) apart, the potential energy between them is (A) \(7.29 \times 10^{-17} \mathrm{~J}\) (B) \(7.88 \times 10^{-9} \mathrm{~J}\) (C) \(7.88 \times 10^{-7} \mathrm{~J}\) (D) \(7.29 \times 10^{-9} \mathrm{~J}\) (E) None of the above
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