Question

Let $ABC$ is an equilateral triangle of side $$10m$$ and $D$ is the midpoint of $BC$ . Charges of $+100,-100\text{ and }+75\mu C$ bs are placed at $B,C,D$ respectively. Find the force on a $+1\mu C$ charge placed at$A$ . (Give your answer in S.I. unit).

Answer 1

Hint:To solve this question firstly we will draw the figure and simplify the forces on $+1\mu C$ . We will calculate the net electric field along the x-axis and then net field of point $A$ , after net force acting on the point charger of $+1\mu C$ .

Formula used:
$E={\displaystyle \frac{kq}{r^2}}$
$\Rightarrow F=qE$
Where, $E$ is the electric field strength,$k=9\times {10}^9$ is the electrostatic constant,$q$ is the charge, $r$ is the distance to the charge creating the field and $F$ is the force.

Complete step by step answer:

Now, we will calculate net field along -axis at point $A$
$$\begin{gathered} E(x)=E(B)\mathrm{Cos}{60}^{\circ }+E(C)\mathrm{Cos}{60}^{\circ } \\ \Rightarrow E(y)=E(D)+E(B)\mathrm{Sin}{60}^{\circ }-E(C)\mathrm{Sin}{60}^{\circ } \end{gathered}$$
Given that,
$a=10m$
$\Rightarrow Q(A)=1\mu C$
$\Rightarrow Q(B)=100\mu C$
$\Rightarrow Q(C)=-100\mu C$
$\Rightarrow Q(D)=75\mu C$
$\Rightarrow k=9\times {10}^{9}N{M}^2{C}^{-2}$
Now electric field at $A$
$$\begin{gathered} \because E_A=k{\displaystyle \frac{Q_A}{a^2}} \\ \Rightarrow E_A=9\times {10}^9\times {\displaystyle \frac{100\times {10}^{-6}}{{10}^2}} \\ \Rightarrow E_A=9000N{C}^{-1} \end{gathered}$$
Now, electric field at $B$ is $E(B)$
$$\begin{gathered} \because E_B=k{\displaystyle \frac{Q_B}{a^2}} \\ \Rightarrow E_B=9\times {10}^9\times {\displaystyle \frac{100\times {10}^{-6}}{{10}^2}} \\ \Rightarrow E_B=9000N{C}^{-1} \end{gathered}$$
Now,electric field at $D$ is $E(D)$
$$\begin{gathered} \because E_D=k{\displaystyle \frac{Q_D}{a^2}} \\ \Rightarrow E_D=9\times {10}^9\times {\displaystyle \frac{75\times {10}^{-6}}{{\displaystyle \frac{3\times {10}^2}{4}}}} \\ \Rightarrow E_D=900\times {10}^{(9-6-2)} \\ \Rightarrow E_D=9000N{C}^{-1} \end{gathered}$$
So,
$$\begin{gathered} E(x)=9000\cos {60}^{\circ }+9000\cos {60}^{\circ } \\ \Rightarrow E(x)=\left(9000\times {\displaystyle \frac{1}{2}}\right)\times 2 \\ \Rightarrow E(x)=9000N{C}^{-1} \end{gathered}$$
Now,
$$\begin{gathered} E(y)=9000+9000\cos {60}^{\circ }+9000\cos {60}^{\circ } \\ \Rightarrow E(y)=9000N{C}^{-1} \end{gathered}$$
So, net electric field at $A$
$$\begin{gathered} E_n=\sqrt{E_x^2+E_y^2} \\ \Rightarrow E_n=\sqrt{{(9000)}^2+{(9000)}^2} \\ \Rightarrow E_n=9000\sqrt{2} \\ \Rightarrow E_n=12727.9N{C}^{-1} \end{gathered}$$
So, net force acting on point charge of $+1\mu C$ is Given By
$$\begin{gathered} F=qE \\ \Rightarrow F=1\times {10}^{-6}\times 12727.9 \\ \therefore F=12.73\times {10}^{-2}N \end{gathered}$$
Hence, the net force acting on $+1\mu C$ is $12.73\times {10}^{-2}N$.

Note:While simplifying the forces don’t make mistakes in angles go step by step calculation to avoid mistakes. The repulsive or attractive interaction between any two charged bodies is called an electric force. Similar to any force, its impact and effects on the given body are described by Newton's laws of motion.