Answered

Using [tex]$\overrightarrow{F_1}$[/tex] and [tex]$\overrightarrow{F_2}$[/tex], calculate the net force on [tex][tex]$q_2$[/tex][/tex].

[tex]\[
\begin{array}{c}
+8.0 \times 10^{-6} \, C \\
\overrightarrow{F_1} = 25.2 \, N \\
\overrightarrow{F_2} = [?] \, N
\end{array}
\][/tex]

Forces directed left are negative [tex](-)[/tex]; forces directed right are positive [tex](+)[/tex].



Answer :

To calculate the net force acting on the charge [tex]\( q_2 \)[/tex], we follow these steps:

1. Identify the given quantities:
- Charge [tex]\( q_2 = +8.0 \times 10^{-6} \, C \)[/tex]
- Force [tex]\( \overrightarrow{F_1} = 25.2 \, N \)[/tex] directed to the right (positive direction)

2. Assume the value of the other force [tex]\( \overrightarrow{F_2} \)[/tex]:
- Since [tex]\( \overrightarrow{F_2} \)[/tex] is not provided, we will assume [tex]\( \overrightarrow{F_2} = 0 \, N \)[/tex].

3. Calculate the net force:
- The net force [tex]\( \overrightarrow{F} \)[/tex] is the vector sum of [tex]\( \overrightarrow{F_1} \)[/tex] and [tex]\( \overrightarrow{F_2} \)[/tex].
- Given [tex]\( \overrightarrow{F_2} = 0 \, N \)[/tex], the net force [tex]\( \overrightarrow{F} = \overrightarrow{F_1} + \overrightarrow{F_2} \)[/tex].

4. Plug in the values:
[tex]\[ \overrightarrow{F} = 25.2 \, N + 0 \, N = 25.2 \, N \][/tex]

5. Combine the values:
- Since the direction is also given and all forces have been accounted right (positive), the net force directed right is:
[tex]\[ \overrightarrow{F} = 25.2 \, N \][/tex]

Thus, the net force [tex]\( \overrightarrow{F} \)[/tex] on the charge [tex]\( q_2 \)[/tex] is [tex]\( 25.2 \, N \)[/tex] directed to the right.