diagram of electrostatic force example
Answers
Explanation:
Example:
Two similar helium filled balloons, each carrying charge Q, are tied to a 30 g weight and are floating at equilibrium as shown in the figure. Find Q.
Solution:
From FBD of mass:
mg=Tcosθ+Tcosθ=2Tcosθ..................................(1)
From FBD of balloon:
F
E
:Electrostatic repulsive force
F E =4πε
0 1
d
2
Q
2
=
4πε
0
1
36
Q
2
Tsinθ=F
E
.........(2)
(2)÷(1),
2
1
tanθ=
mg
F
E
=
8
3
⇒F
E
=
8
3
mg (∵tanθ=
4
3
)
∴
4πε
0
1
36
Q
2
=
8
3
×30×10
−3
×10
Q
2
=4.5×10
−10
Q=21.2μC
Answer:
Electrostatic force between three non-colinear point charges
\begin{figure*}
\epsfysize =2in
\centerline{\epsffile{fig1.eps}}
\end{figure*}
Question: Suppose that three point charges, $q_a$, $q_b$, and $q_c$, are arranged at the vertices of a right-angled triangle, as shown in the diagram. What is the magnitude and direction of the electrostatic force acting on the third charge if $q_a=-6.0\,\mu{\rm C}$, $q_b=+4.0\,\mu{\rm C}$, $q_c = +2.0\,\mu{\rm C}$, $a=4.0$m, and $b=3.0$m?
Solution: The magnitude $f_{ac}$ of the force ${\bf f}_{ac}$ exerted by charge $q_a$ on charge $q_c$ is given by
\begin{displaymath}
f_{ac} = k_e\,\frac{\vert q_a\vert\, q_c}{c^2}= (8.988\times...
...2\times 10^{-6})}
{(4^2+3^2)} = 4.31 \times 10^{-3} \,{\rm N},
\end{displaymath}
where use has been made of the Pythagorean theorem. The force is attractive (since charges $q_a$ and $q_c$ are of opposite sign). Hence, the force is directed from charge $q_c$ towards charge $q_a$, as shown in the diagram. The magnitude $f_{bc}$ of the force ${\bf f}_{bc}$ exerted by charge $q_b$ on charge $q_c$ is given by
\begin{displaymath}
f_{bc} = k_e\,\frac{q_b q_c}{b^2}= (8.988\times 10^9)\,\frac...
...)\,(2\times 10^{-6})}
{(3^2)} = 7.99 \times 10^{-3} \,{\rm N}.
\end{displaymath}
The force is repulsive (since charges $q_b$ and $q_c$ are of the same sign). Hence, the force is directed from charge $q_b$ towards charge $q_c$, as shown in the diagram. Now, the net force acting on charge $q_c$ is the sum of ${\bf f}_{ac}$ and ${\bf f}_{bc}$. Unfortunately, since ${\bf f}_{ab}$ and ${\bf f}_{bc}$ are vectors pointing in different directions, they cannot be added together algebraically. Fortunately, however, their components along the $x$- and $y$-axes can be added algebraically. Now, it is clear, from the diagram, that ${\bf f}_{bc}$ is directed along the $+x$-axis. If follows that
$\displaystyle f_{bc\,x}$ $\textstyle =$ $\displaystyle f_{bc} = 7.99 \times 10^{-3} \,{\rm N},$
$\displaystyle f_{bc\,y}$ $\textstyle =$ $\displaystyle 0.$
It is also clear, from the diagram, that ${\bf f}_{ac}$ subtends an angle
\begin{displaymath}
\theta =\tan^{-1} (a/b) = \tan^{-1}(4/3) = 53.1^\circ
\end{displaymath}
with the $-x$-axis, and an angle $90^\circ-\theta$ with the $+y$-axis. It follows from the conventional laws of vector projection that
$\displaystyle f_{ac\,x}$ $\textstyle =$ $\displaystyle -f_{ac}\,\cos\theta = -(4.31 \times 10^{-3})\,(0.6)=-2.59\times 10^{-3}\,{\rm N},$
$\displaystyle f_{ac\,y}$ $\textstyle =$ $\displaystyle f_{ac}\,\cos(90^\circ -\theta) = f_{ac} \,\sin\theta = (4.31 \times 10^{-3})\,(0.8) = 3.45 \times 10^{-3}\,{\rm N}.$
The $x$- and $y$-components of the resultant force ${\bf f}_c$ acting on charge $q_c$ are given by
$\displaystyle f_{c\,x}$ $\textstyle =$ $\displaystyle f_{ac\,x} + f_{bc\,x} = -2.59\times 10^{-3} + 7.99 \times 10^{-3} =5.40\times
10^{-3}\,{\rm N},$
$\displaystyle f_{c\,y}$ $\textstyle =$ $\displaystyle f_{ac\,y} + f_{bc\,y} = 3.45 \times 10^{-3}\,{\rm N}.$
Thus, from the Pythagorean theorem, the magnitude of the resultant force is
\begin{displaymath}
f_c = \sqrt{(f_{c\,x})^2 + (f_{c\,y})^2} = 6.4\times 10^{-3}\,{\rm N}.
\end{displaymath}
Furthermore, the resultant force subtends an angle
\begin{displaymath}
\phi = \tan^{-1} (f_{c\,y}/f_{c\,x}) = 32.6^\circ
\end{displaymath}
with the $+x$-axis, and an angle $90^\circ-\phi = 57.4^\circ$ with the $+y$-axis.