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Chapter 16: Problem 34

Positive point charges \(q\) and \(2 q\) are located at \(x=0\) and \(x=3 d,\) respectively. Are there any points not on the \(x\) -axis where $\overrightarrow{\mathbf{E}}=0 ?$ Explain.

Short Answer

Expert verified
Based on the above solution, the analysis and equations show that there are no points not on the x-axis where the electric field is equal to zero. This conclusion is drawn from equation (5) which demonstrates that it is independent of θ1 and θ2 and thus does not satisfy the condition of having y≠0.

Step by step solution

01

Electric field due to individual charges

The electric field at a point P due to a single point charge q is given by \(\vec{E} = \dfrac{k*q}{r^2}\hat{r}\), where \(k\) is the electrostatic constant, \(q\) is the charge, \(r\) is the distance between the charge and the point P, and \(\hat{r}\) is the unit vector pointing from the charge to P.
02

Components of electric fields

Let's consider a point P in the x-y plane that is not on the x-axis (i.e., \(y\neq0\)). Let \(r_1\) and \(r_2\) denote the distances from the point P to the charges \(q\) and \(2q\), respectively. We will find the components of the electric fields \(\vec{E}_1\) and \(\vec{E}_2\) due to these two charges along the x and y directions. Using trigonometric functions, we have, For \(\vec{E}_1\): \(E_{1x} = \dfrac{k*q}{r_1^2}\cos{\theta_1}\) and \(E_{1y} = \dfrac{k*q}{r_1^2}\sin{\theta_1}\) For \(\vec{E}_2\): \(E_{2x} = -\dfrac{k*2q}{r_2^2}\cos{\theta_2}\) and \(E_{2y} = \dfrac{k*2q}{r_2^2}\sin{\theta_2}\) The negative sign in \(E_{2x}\) is because its x-component is directed opposite to the \(x\)-axis.
03

Equating the x and y components of the electric fields

To find the point P where the total electric field becomes zero, we first equate both the x-components and then the y-components of the electric fields: \(E_{1x} + E_{2x} = 0 \rightarrow \dfrac{k*q}{r_1^2}\cos{\theta_1} + \dfrac{-k*2q}{r_2^2}\cos{\theta_2}=0\) (1) \(E_{1y} + E_{2y} = 0 \rightarrow \dfrac{k*q}{r_1^2}\sin{\theta_1} + \dfrac{k*2q}{r_2^2}\sin{\theta_2}=0\) (2)
04

Analyzing the equations

Dividing both sides of equation (1) by \(k*q\) and equation (2) by \(k*2q\), we get: \(\dfrac{1}{r_1^2}\cos{\theta_1} = \dfrac{2}{r_2^2}\cos{\theta_2}\) (3) \(\dfrac{1}{r_1^2}\sin{\theta_1} = \dfrac{1}{r_2^2}\sin{\theta_2}\) (4) Now, squaring both equations and adding them, we have: \(\dfrac{1}{r_1^4}\cos^2{\theta_1} + \dfrac{2}{r_1^4r_2^2}\cos{\theta_1}\cos{\theta_2} + \dfrac{4}{r_2^4}\cos^2{\theta_2} = \dfrac{1}{r_1^4}\sin^2{\theta_1} + \dfrac{1}{r_1^2r_2^2}\sin{\theta_1}\sin{\theta_2} + \dfrac{1}{r_2^4}\sin^2{\theta_2}\) Using the identity \(\sin^2{\theta}+\cos^2{\theta}=1\), we get: \(\dfrac{1}{r_1^4} + \dfrac{4}{r_2^4} = \dfrac{3}{r_1^2r_2^2}\) (5)
05

Conclusion

Looking at equation (5), we observe that it is independent of \(\theta_1\) and \(\theta_2\). As a result, we cannot find any solution for this equation that has \(y\neq0\). In other words, there is no point not on the \(x\)-axis where the electric field \(\vec{E}=0\).

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