Question

Four charges of \[6\times {{10}^{-8}}C\] coulomb each are placed at the corner of a square whose sides are 3 cm each. calculate the resultant force on each charger and show in direction and a diagram drawn not to be scale

Answer 1

Hint: We will calculate only for one of the charges because the magnitude of force will be equal for all the cases for all the four charges. We had represented them in the figure. Since force is a vector quantity, we add it using either triangle law of vector addition or the parallelogram law of vector addition.

Complete step by step answer:Magnitude of each charge is q=\[6\times {{10}^{-8}}C\]
Side of square is, r= 3 cm = 0.03 m
Magnitude of diagonal= \[\sqrt{{{r}^{2}}+{{r}^{2}}}=\sqrt{2}r\]
Force on A due to B:
From coulomb’s law \[F=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\], where k is a constant and r is the separation between the two charges.
\[{{F}_{BA}}=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}=\dfrac{k{{q}^{2}}}{{{r}^{2}}}\]
Force on A due to C:
\[{{F}_{CA}}=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}=\dfrac{k{{q}^{2}}}{2{{r}^{2}}}\]
Force on A due to D:
\[{{F}_{DA}}=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}=\dfrac{k{{q}^{2}}}{{{r}^{2}}}\]
Now from the figure it is clear that the force on A due to B and D are at perpendicular so resultant force of them will point in the direction of force due to C. Thus, resultant force will be,
\[\begin{align}
  & F={{F}_{BA}}+{{F}_{CA}}+{{F}_{DA}} \\
 & F={{F}_{CA}}+({{F}_{BA}}+{{F}_{DA}}) \\
\end{align}\]
Putting the values,
\[\begin{align}
  & F=\dfrac{k{{q}^{2}}}{2{{r}^{2}}}+\sqrt{{{\{\dfrac{k{{q}^{2}}}{{{r}^{2}}}\}}^{2}}+{{\{\dfrac{k{{q}^{2}}}{{{r}^{2}}}\}}^{2}}} \\
 & \Rightarrow F=\dfrac{k{{q}^{2}}}{2{{r}^{2}}}+\dfrac{k{{q}^{2}}}{{{r}^{2}}}\times \sqrt{2} \\
 & \Rightarrow F=\dfrac{k{{q}^{2}}}{{{r}^{2}}}\{\dfrac{1}{2}+\sqrt{2}\} \\
\end{align}\]
Now we put the values of k, q and r, we get,
\[\begin{align}
  & F=\dfrac{9\times {{10}^{9}}\times {{(6\times {{10}^{-8}})}^{2}}}{{{(3\times {{10}^{-2}})}^{2}}}\{\dfrac{1}{2}+\sqrt{2}\} \\
 & \Rightarrow F=0.0689N \\
\end{align}\]

Note:We have taken the magnitudes and direction is shown in the figure. Since force is a vector quantity. So, the magnitude of net force on any charge is 0.0689 N. Here k is a constant whose value is \[9\times {{10}^{9}}N{{m}^{2}}/{{C}^{2}}\]