ANSWER
A) 5 N
EXPLANATION
Given:
• Charge q₁ = -4 x 10⁻⁷ C
,
• The position of q₁, (3, 0)
,
• Charge q₂ = 12 x 10⁻⁷ C
,
• The position of q₂, (0, 6)
,
• Charge q₃ = 10⁻⁶ C
,
• The position of q₃, (0, 0)
Find:
• The intensity of the resulting force acting on q₃, F
Charges q₁ and q₃ have opposite charges, so q₁ exerts an attraction force on q₃. On the other hand, q₂ has the same sign as q₃, so q₂ exerts a repulsion force on q₃. So, we have,
The magnitude of the force exerted by q₁ is given by Coulomb's law,
[tex]F_1=k\cdot\frac{q_1q_3}{r_1^2}[/tex]
Where r₁ is the distance between charges q₁ and q₃ in meters,
[tex]r_1=3cm=0.03m[/tex]
So, the force F₁ is,
[tex]F_1=9\cdot10^9Nm^2/C^2\cdot\frac{4\cdot10^{-7}C\cdot1\cdot10^{-6}C}{0.03^2m^2}=4N[/tex]
Similarly, the force exerted by q₂ on q₃ is given by,
[tex]F_2=k\cdot\frac{q_2q_3}{r_2^2}[/tex]
Where r₂ is the distance between q₂ and q₃ in meters,
[tex]r_2=6cm=0.06m[/tex]
So, the force F₂ is,
[tex]F_2=9\cdot10^9Nm^2/C^2\cdot\frac{12\cdot10^{-7}C\cdot1\cdot10^{-6}C}{0.06^2m^2}=3N[/tex]
As illustrated in the diagram above, these two forces are perpendicular, so the resultant is the hypotenuse of the right triangle they form. By the Pythagorean Theorem,
[tex]F=\sqrt{F_1^2+F_2^2}=\sqrt{4^2N^2+3^2N^2}=\sqrt{16N^2+9N^2}=\sqrt{25N^2}=5N[/tex]
Hence, the resulting force's intensity is 5 N.
