Answer:
Assumption: there is no object between this point charge and the observer.
Explanation:
The electric field of a point charge is inversely proportional to the square of the distance from that point charge.
Let [tex]k[/tex] denote Coulomb's constant ([tex]k \approx 8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-1}[/tex].) Let the magnitude of that point charge be [tex]q[/tex]. At a distance of [tex]r[/tex] from this charge, the electric field due to this charge would be:
[tex]\displaystyle E = \frac{k \cdot q}{r^{2}}[/tex].
Convert the magnitude of the point charge in this question to standard units:
[tex]q = 1\; \rm nC = 10^{-9}\; \rm C[/tex].
Apply that equation to find the magnitude of the electric field due to this point charge:
[tex]r = 1\; \rm m[/tex]:
[tex]\begin{aligned} E &= \frac{k \cdot q}{r^{2}} \\ &= \frac{8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2} \times 10^{-9}\; \rm C}{(1\; \rm m)^{2}} \\ &\approx 9.0\; \rm N \cdot C^{-1}\end{aligned}[/tex].
[tex]r = 2\; \rm m[/tex]:
[tex]\begin{aligned} E &= \frac{k \cdot q}{r^{2}} \\ &= \frac{8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2} \times 10^{-9}\; \rm C}{(2\; \rm m)^{2}} \\ &\approx 2.2\; \rm N \cdot C^{-1}\end{aligned}[/tex].
[tex]r = 3\; \rm m[/tex]:
[tex]\begin{aligned} E &= \frac{k \cdot q}{r^{2}} \\ &= \frac{8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2} \times 10^{-9}\; \rm C}{(3\; \rm m)^{2}} \\ &\approx 1.0\; \rm N \cdot C^{-1}\end{aligned}[/tex].
The direction of the electric field at a point is the same as the direction of a force from this field onto a positive point charge at this point.
Because the [tex](+1\; \rm nC)[/tex] point charge here is positive, the electric field of this charge would repel other positive point charges. Hence, the electric field around this [tex](+1\; \rm nC)\![/tex] point charge at any point in the field would point away from this charge.
