Answer :
Answer:
Approximately [tex]3.8\times 10^{7}\; {\rm N\cdot C^{-1}}[/tex].
Explanation:
The electric field at a given point is a vector quantity. Let [tex]k[/tex] denote Coulomb's Constant. At a distance of [tex]r[/tex] from a point charge of magnitude [tex]q[/tex], the magnitude of the electric field from that point charge would be:
[tex]\displaystyle \frac{k\, q}{r^{2}}[/tex].
The direction of the electric field is the same as the direction of the electrostatic force that a positive test charge [tex]q_\text{test}[/tex] would experience:
- If the point charge [tex]q[/tex] is positive, [tex]q[/tex] would repel [tex]q_\text{test}[/tex], which is also positively charged, and the resultant electric field would point away from this point charge.
- Conversely, if the point charge [tex]q[/tex] is negative, [tex]q[/tex] would attract [tex]q_\text{test}[/tex], and the resultant electric field would point towards this point charge.
When there are two point charges near a given position, the resultant electric field would be the vector sum of the electric field from each charge.
In this question, both the [tex](-5.6)\; {\rm \mu C}[/tex] charge and the [tex](+5.0)\; {\rm \mu C}[/tex] charge are at a distance of [tex](10\; {\rm cm}) / 2 = 5\; {\rm cm}[/tex] from the given position "midway" between the two point charges.
For the [tex](-5.6)\; {\rm \mu C}[/tex] charge, magnitude of electrostatic charge would be [tex]q = 5.6\; {\rm \mu C} = 5.6 \times 10^{-6}\; {\rm C}[/tex]. At a distance of [tex]r = 5\; {\rm cm} = 5 \times 10^{-2}\; {\rm m}[/tex], magnitude of the electric field from this charge would be:
[tex]\begin{aligned} E &= \frac{k\, q}{r^{2}} \\ &\approx \frac{(8.99 \times 10^{9}\; {\rm N\cdot m^{2}\cdot C^{-2}})\times (5.6 \times 10^{-6}\; {\rm C})}{\left(5 \times 10^{-2}\; {\rm m}\right)^{2}} \\ &\approx 2.01 \times 10^{7}\; {\rm N \cdot C^{-1}} \end{aligned}[/tex].
Because this point charge is negative, this charge would attract any positive test charge placed at the given position. Hence, the electric field from this point charge would point towards this [tex](-5.6)\; {\rm \mu C}[/tex] charge from the midpoint.
Similarly, for the [tex](+5.0)\; {\rm \mu C}[/tex] charge, magnitude of electrostatic charge would be [tex]q = 5.0\; {\rm \mu C} = 5.0 \times 10^{-6}\; {\rm C}[/tex]. At a distance of [tex]r = 5\; {\rm cm} = 5 \times 10^{-2}\; {\rm m}[/tex], magnitude of the electric field from this charge would be:
[tex]\begin{aligned} E &= \frac{k\, q}{r^{2}} \\ &\approx \frac{(8.99 \times 10^{9}\; {\rm N\cdot m\cdot C^{-2}})\times (5.0\times 10^{-6}\; {\rm C})}{\left(5 \times 10^{-2}\; {\rm m}\right)^{2}} \\ &\approx 1.80 \times 10^{7}\; {\rm N \cdot C^{-1}} \end{aligned}[/tex].
Because this point charge is positive, this charge would repel any positive test charge placed at the given position. Hence, the electric field from this point charge would point away from this [tex](+5.0)\; {\rm \mu C}[/tex] charge from the midpoint, which would be towards the [tex](-5.6)\; {\rm \mu C}[/tex] charge.
In other words, at the midpoint, the electric field from the [tex](-5.6)\; {\rm \mu C}[/tex] charge and the [tex](+5.0)\; {\rm \mu C}[/tex] charge would both point in the same direction towards the [tex](-5.6)\; {\rm \mu C}[/tex] charge. Since the resultant electric field at the midpoint is the vector sum of the contribution from each point charge, magnitude of the resultant electric field would be the sum of the magnitude of the electric field from each point charge:
[tex]2.01 \times 10^{7}\; {\rm N \cdot C^{-1}} + 1.80 \times 10^{7}\; {\rm N\cdot C^{-1}} \approx 3.8\times 10^{7}\; {\rm N\cdot C^{-1}}[/tex].
