What is the force on the 1.0nC charge in the figure? Give the magnitude
a.) What is the force of the 1.0 nC in the figure (in Newtons)
b.) Give the direction (degrees above horizontal)
Homework Answers
The force acting on \(1 \mathrm{nC}\) due to \(2 \mathrm{nC}\) separated by
\(1 \mathrm{~cm}\) is given by
$$ \begin{aligned} F_{1}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}} &=9 \times 10^{9} \frac{\left(1 \times 10^{-9}\right)\left(2 \times 10^{-9}\right)}{(0.01)^{2}} \\ &=18 \times 10^{-5} \mathrm{~N} \end{aligned} $$
And this force is repulsive.
Resolving the force \(F_{1}\) into components
The horizontal component is
$$ \mathrm{F}_{1 x}=\mathrm{F}_{1} \cos \theta=\left(18 \times 10^{-5} \mathrm{~N}\right) \cos \left(60^{0}\right)=9 \times 10^{-5} \mathrm{~N} $$
And the vertical component is
$$ F_{1 y}=F_{1} \sin \theta=\left(18 \times 10^{-5} \mathrm{~N}\right) \sin \left(60^{\circ}\right)=15.58 \times 10^{-5} \mathrm{~N} $$
Similarly
The force acting on \(1 \mathrm{nC}\) due to \(-2 \mathrm{nC}\) separated by
\(1 \mathrm{~cm}\) is given by
$$ \begin{aligned} F_{2}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}} &=9 \times 10^{9} \frac{\left(1 \times 10^{-9}\right)\left(2 \times 10^{-9}\right)}{(0.01)^{2}} \\ &=18 \times 10^{-5} \mathrm{~N} \end{aligned} $$
And this force is attractive.
Resolving the force \(\mathrm{F}_{2}\) into components
The horizontal component is
$$ F_{2 x}=F_{2} \cos \theta=\left(18 \times 10^{-5} \mathrm{~N}\right) \cos \left(-60^{\circ}\right)=9 \times 10^{-5} \mathrm{~N} $$
And the vertical component is
$$ \mathrm{F}_{2 y}=\mathrm{F}_{2} \sin \theta=\left(18 \times 10^{-5} \mathrm{~N}\right) \sin \left(-60^{\circ}\right)=-15.58 \times 10^{-5} \mathrm{~N} $$
The net force acting along horizontal direction is
$$ \mathrm{F}_{x}=\mathrm{F}_{1 x}+\mathrm{F}_{2 x}=9 \times 10^{-5} \mathrm{~N}+9 \times 10^{-5} \mathrm{~N}=18 \times 10^{-5} \mathrm{~N} $$
The net force acting along vertical direction is
$$ \mathrm{F}_{y}=\mathrm{F}_{2 y}+\mathrm{F}_{2 y}=15.58 \times 10^{-5} \mathrm{~N}-15.58 \times 10^{-5} \mathrm{~N}=0 $$
Hence the net force acting on \(1 \mathrm{nC}\) charge is
\(F=18 \times 10^{-5} \mathrm{~N}\)
and the direction it makes with the \(\mathrm{x}\) - axis is \(0^{0}\)
In vector form \(F=18 \times 10^{-5} \hat{i}\).
F1 = 1.8*10^-4 cos60 + 1.8*10^-4 sin60
F3 = 1.8*10^-4 cos60 - 1.8*10^-4 sin60
Fnet = F1+F3 = 1.8 * 10 ^-4 N
The direction of the force is to the right
Add Answer to:
What is the force on the 1.0nC charge in the figure? Give the magnitude
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> Did you perhaps forget a decimal point between 1 and 8? It should be 1.8 and should also be 10^-4 instead of 10^-5
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