The least number of component of a vector quantity is two. These are the x-component and the y-component.
The resultant vector, or vector as we refer to it in this item, can be calculated through the equation,
RV = sqrt ((Vx)² + (Vy)²)
From the equation, it can be noted that if we let Vx equal to zero,
RV = Vy
Similarly, if we let Vy be equal to zero then,
RV = Vx
Thus, it is still possible for the vector to become nonzero even if one of its components is zero.
Complete question:
if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.
Answer:
The mutual force between the two point charges is 319.64 N
Explanation:
Given;
distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m
value of the charges, q₁ and q₂ = 2 μC and - μ4 C
Apply Coulomb's law;

where;
F is the force of attraction between the two charges
|q₁| and |q₂| are the magnitude of the two charges
r is the distance between the two charges
k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

Therefore, the mutual force between the two point charges is 319.64 N
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