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

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

Step-by-step 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

`(k*q_(1) )/(d_(1)^(2)  ) +(k*q_(2) )/(d_(2)^(2)  ) -(k*q_(3) )/(d_(3)^(2)  ) =0`

We eliminate k

`(q_(1) )/(d_(1) ^(2) ) +(q_(2) )/(d_(2) ^(2) )+(q_(3) )/(d_(3) ^(2) )=0`

We replace data

`(3*10^(-6) )/((3*10^(-2))^(2) ) +(10*10^(-6) )/((4*10^(-2))^(2) ) +(6*10^(-6) )/(d_(3) ^(2) ) =0`

we eliminate 10⁻⁶

`(3)/(9*10^(-4) ) +(10)/(16*10^(-4) ) =(6)/(d_(3)^(2)  )`

`((1)/(10^(-4) )) *((1)/(3) +(5)/(8)) =(6)/(d_(3)^(2)  )`

`(23*10^(4) )/(24) =(6)/(d_(3) ^(2) )`

`d_(3) =\sqrt{(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