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Chapter 22: Problem 25

\( \mathrm{~A}+48.00-\mathrm{nC}\) point charge is placed on the \(x\) -axis at \(x=4.000 \mathrm{~m},\) and \(\mathrm{a}-24.00-\mathrm{n} \mathrm{C}\) point charge is placed on the \(y\) -axis at \(y=-6.000 \mathrm{~m} .\) What is the direction of the electric field at the origin?

Short Answer

Expert verified
Answer: The direction of the net electric field at the origin can be found using the formula \(\theta = \arctan(\frac{E_{nety}}{E_{netx}}) = \arctan(\frac{E_B}{-E_A})\), where \(E_A\) and \(E_B\) are the electric fields due to charges A and B, respectively. Plug in the values of A and B, calculate \(E_A\) and \(E_B\), and then find the angle \(\theta\) to determine the direction of the net electric field at the origin.

Step by step solution

01

Finding the electric field due to charge A at the origin

To find the electric field due to the first charge A at the origin, use the formula for the electric field from a point charge: \(E = k* \frac{Q}{r^2}\), where \(k = 8.9875517923*10^9 N m^2/C^2\) is Coulomb's constant, \(Q\) is the charge, and \(r\) is the distance between the charge and the point at which we want to find the electric field. The distance between charge A and the origin is \(x=4m\). Therefore, the electric field at the origin due to charge A is: \(E_A = k* \frac{A}{(4m)^2}\).
02

Finding the electric field due to charge B at the origin

Repeat the process for charge B, which is placed on the y-axis at \(y=-6m\). The distance between charge B and the origin is \(6m\). Therefore, the electric field at the origin due to charge B is: \(E_B = k* \frac{B}{(6m)^2}\), where B is -24.00 nC.
03

Determining the components of the electric fields

We know that the electric field due to charge A is in the negative x-direction and the electric field due to charge B is in the positive y-direction. Thus, the components of the electric fields are: \(E_{Ax}=-E_A, E_{Ay}=0\) \(E_{Bx}=0, E_{By}=E_B\)
04

Vector addition of the electric fields

Now, we'll find the net electric field at the origin by adding the components of the electric fields due to charges A and B: \(E_{netx} = E_{Ax} + E_{Bx} = -E_A\) \(E_{nety} = E_{Ay} + E_{By} = E_B\)
05

Calculating the direction of the net electric field

To find the direction of the net electric field at the origin, we need to determine the angle with respect to the x-axis. We use the arctangent function to find the angle: \(\theta = \arctan(\frac{E_{nety}}{E_{netx}}) = \arctan(\frac{E_B}{-E_A})\) Now, you can plug in the values of A, B and find \(E_A\) and \(E_B\) to calculate the angle, \(\theta\), which gives the direction of the net electric field at the origin.

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Most popular questions from this chapter

Two uniformly charged insulating rods are bent in a semicircular shape with radius \(r=10.0 \mathrm{~cm} .\) If they are positioned so they form a circle but do not touch and have opposite charges of \(+1.00 \mu \mathrm{C}\) and \(-1.00 \mu \mathrm{C}\) find the magnitude and direction of the electric field at the center of the composite circular charge configuration.

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