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hastings
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[SOLVED] finding coulmb's force
Problem:
A very long straight wire has a uniform linear density charge, λ (every dx, linear unit, has a λ charge). At a distance D, from the wire, there's a rectangular layer with length=a and height=b and a uniform superficial density charge, σ (every dS, superficial unit, has a σ charge). Find the electrostatic force between the 2 objects.
What I did:
(1)[tex] F=\int dF[/tex]
(2)[tex]dF=\frac{1}{4\pi \varepsilon_0} \frac{dq_1 \cdot dq_2}{x^2}[/tex]
(3)[tex] dq_1=\lambda dx[/tex]
(4)[tex] dq_2=\sigma dS \text{ (* When integrating dS becomes S=a $\cdot$ b)}[/tex]
(5)[tex]F=\int dF=\int_D^{D+a} \frac{1}{4\pi \varepsilon_0} \frac{(\lambda dx) \cdot \sigma dS }{x^2} = \frac{\lambda \sigma (a \cdot b)}{4\pi \varepsilon_0} \int_D^{D+a} \frac{1}{x^2}dx[/tex]
(6)[tex]=\frac{\lambda \sigma (a \cdot b)}{4\pi \varepsilon_0}(-\frac{1}{x})_D^{D+a}= \frac{\lambda \sigma (a \cdot b)}{4\pi \varepsilon_0} (\frac{1}{D}-\frac{1}{D+a}) =\frac{\lambda \sigma (a \cdot b)}{4\pi \varepsilon_0}\frac{a}{D(D+a)}[/tex]
(7)[tex]=\frac{\lambda \sigma (a^2 b)}{4\pi \varepsilon_0 \ D(D+a)} [/tex]
Is my reasoning right? Or, where did I mistake? Please check out the little drawing in the attachment.
Thank you.
Problem:
A very long straight wire has a uniform linear density charge, λ (every dx, linear unit, has a λ charge). At a distance D, from the wire, there's a rectangular layer with length=a and height=b and a uniform superficial density charge, σ (every dS, superficial unit, has a σ charge). Find the electrostatic force between the 2 objects.
What I did:
(1)[tex] F=\int dF[/tex]
(2)[tex]dF=\frac{1}{4\pi \varepsilon_0} \frac{dq_1 \cdot dq_2}{x^2}[/tex]
(3)[tex] dq_1=\lambda dx[/tex]
(4)[tex] dq_2=\sigma dS \text{ (* When integrating dS becomes S=a $\cdot$ b)}[/tex]
(5)[tex]F=\int dF=\int_D^{D+a} \frac{1}{4\pi \varepsilon_0} \frac{(\lambda dx) \cdot \sigma dS }{x^2} = \frac{\lambda \sigma (a \cdot b)}{4\pi \varepsilon_0} \int_D^{D+a} \frac{1}{x^2}dx[/tex]
(6)[tex]=\frac{\lambda \sigma (a \cdot b)}{4\pi \varepsilon_0}(-\frac{1}{x})_D^{D+a}= \frac{\lambda \sigma (a \cdot b)}{4\pi \varepsilon_0} (\frac{1}{D}-\frac{1}{D+a}) =\frac{\lambda \sigma (a \cdot b)}{4\pi \varepsilon_0}\frac{a}{D(D+a)}[/tex]
(7)[tex]=\frac{\lambda \sigma (a^2 b)}{4\pi \varepsilon_0 \ D(D+a)} [/tex]
Is my reasoning right? Or, where did I mistake? Please check out the little drawing in the attachment.
Thank you.
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