Answer:
The charge q₃ must be placed at X = +2.5 cm
Explanation:
Conceptual analysis
The electric field at a point P due to a point charge is calculated as follows:
E = k*q/d²
E: Electric field in N/C
q: charge in Newtons (N)
k: electric constant in N*m²/C²
d: distance from charge q to point P in meters (m)
The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.
Equivalences
1µC= 10⁻6 C
1cm= 10⁻² m
Data
k = 8.99*10⁹ N×m²/C²
q₁ =+3 µC =3*10⁻⁶ C
q₂ = -10 µC =-10*10⁻⁶ C
q₃= +6µC =+6*10⁻⁶ C
d₁ = 3cm =3×10⁻² m
d₂ = 4cm = 4×10⁻² m
Graphic attached
The attached graph shows the field due to the charges:
E₁:Field at point P due to charge q₁. As the charge is positive ,the field leaves the charge. The direction of E1 is (+ x).
E₂: Field at point P due to charge q₂. As the charge is positive ,the field leaves the charge. The direction of E1 is (+ x).
Problem development
E₃: Field at point P due to charge q₃. As the charge q₃ is positive, the field leaves the charge.
The direction of E₃ must be (- x) so that the electric field can be equal to zero at point P since E₁ and E₂ are positive, then, q₃must be located to the right of point P.
We make the algebraic sum of fields at point P due to the charges q1, q2, and q3:
E₁+E₂-E₃=0
[tex]\frac{k*q_{1} }{d_{1}^{2} } +\frac{k*q_{2} }{d_{2}^{2} } -\frac{k*q_{3} }{d_{3}^{2} } =0[/tex]
We eliminate k
[tex]\frac{q_{1} }{d_{1} ^{2} } +\frac{q_{2} }{d_{2} ^{2} }+\frac{q_{3} }{d_{3} ^{2} }=0[/tex]
We replace data
[tex]\frac{3*10^{-6} }{(3*10^{-2})^{2} } +\frac{10*10^{-6} }{(4*10^{-2})^{2} } +\frac{6*10^{-6} }{d_{3} ^{2} } =0[/tex]
we eliminate 10⁻⁶
[tex]\frac{3}{9*10^{-4} } +\frac{10}{16*10^{-4} } =\frac{6}{d_{3}^{2} }[/tex]
[tex](\frac{1}{10^{-4} }) *(\frac{1}{3} +\frac{5}{8}) =\frac{6}{d_{3}^{2} }[/tex]
[tex]\frac{23*10^{4} }{24} =\frac{6}{d_{3} ^{2} }[/tex]
[tex]d_{3} =\sqrt{\frac{6*24}{23*10^{4} } }[/tex]
[tex]d_{3} =2.5*10^{-2} m\\d_{3} =2.5 cm[/tex]
The charge q₃ must be placed at X = +2.5 cm
