Question

A point charge of 3 µC is located at x = -3.0 cm, and a second point charge of -10 µC is located at x = +4.0 cm. Where should a third charge of +6.0 µC be placed so that the electric field at x = 0 is zero?

Answer 1

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

$\frac{k*q_{1} }{d_{1}^{2} } +\frac{k*q_{2} }{d_{2}^{2} } -\frac{k*q_{3} }{d_{3}^{2} } =0$

We eliminate k

$\frac{q_{1} }{d_{1} ^{2} } +\frac{q_{2} }{d_{2} ^{2} }+\frac{q_{3} }{d_{3} ^{2} }=0$

We replace data

$\frac{3*10^{-6} }{(3*10^{-2})^{2} } +\frac{10*10^{-6} }{(4*10^{-2})^{2} } +\frac{6*10^{-6} }{d_{3} ^{2} } =0$

we eliminate 10⁻⁶

$\frac{3}{9*10^{-4} } +\frac{10}{16*10^{-4} } =\frac{6}{d_{3}^{2} }$

$(\frac{1}{10^{-4} }) *(\frac{1}{3} +\frac{5}{8}) =\frac{6}{d_{3}^{2} }$

$\frac{23*10^{4} }{24} =\frac{6}{d_{3} ^{2} }$

$d_{3} =\sqrt{\frac{6*24}{23*10^{4} } }$

$d_{3} =2.5*10^{-2} m\\d_{3} =2.5 cm$

The charge q₃ must be placed at X = +2.5 cm