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Three charges of each magnitude \(100 \mu \mathrm{C}\) are placed at the corners \(A, B\) and \(C\) of an equilateral triangle of side \(4 \mathrm{~m}\). If the charges at points \(A\) and \(C\) are positive and the charge at point \(B\) is negative, then the magnitude of total force acting on the charge at \(C\) and angle made by it with \(A C\) are
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Verified Answer
The correct answer is:
\(5.625 \mathrm{~N}, 60^{\circ}\)
According to the question, we can draw the following diagram,
From the question, it clear that the charge on the each corner is \(100 \mu \mathrm{C}\).
So,
\(\begin{aligned}
F_{\text {net }} & =F \\
& =\frac{k Q_1 Q_2}{r^2}
\end{aligned}\)
Given, \(Q_1=Q_2=100 \mu \mathrm{C}=100 \times 10^{-6} \mathrm{C}, \quad\left[1 \mu \mathrm{C}=10^{-6} \mathrm{C}\right]\)
and \(r=4 \mathrm{~m}\)
Putting the given values in Eq. (i), we get
\(\begin{aligned}
& =\frac{9 \times 10^9 \times\left(100 \times 10^{-6}\right)^2}{(4)^2} \\
F_{\text {net }} & =5.625 \mathrm{~N}, 60^{\circ}
\end{aligned}\)
From the question, it clear that the charge on the each corner is \(100 \mu \mathrm{C}\).
So,
\(\begin{aligned}
F_{\text {net }} & =F \\
& =\frac{k Q_1 Q_2}{r^2}
\end{aligned}\)
Given, \(Q_1=Q_2=100 \mu \mathrm{C}=100 \times 10^{-6} \mathrm{C}, \quad\left[1 \mu \mathrm{C}=10^{-6} \mathrm{C}\right]\)
and \(r=4 \mathrm{~m}\)
Putting the given values in Eq. (i), we get
\(\begin{aligned}
& =\frac{9 \times 10^9 \times\left(100 \times 10^{-6}\right)^2}{(4)^2} \\
F_{\text {net }} & =5.625 \mathrm{~N}, 60^{\circ}
\end{aligned}\)
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