Question

A -5.40nC point charge is on the x axis at x = 1.25m . A second point charge Q is on the x axis at -0.625m.

A) What must be the charge Q for the resultant electric field at the origin to be 50.0N/C in the +x direction?
B) What must be the charge Q for the resultant electric field at the origin to be 50.0N/C in the -x direction?

Answer 1

Answe

a) Q = 0.820 10⁻⁹ C , b) Q = -3.52 10⁻⁹ C

Step-by-step explanation:

The electric field is given by the formula

E = k q / r²

where E is a vector quantity, so it must be added as a vector

E_total = E₁ + E₂

let's look for the two electric fields

E₁ = k q₁ / r₁²

E₁ = 9 10⁹ 5.4 10⁻⁹ / 1.25²

E₁ = 31.10 N / C

E2 = k Q / r₂²

E2 = 9 10⁹ Q / 0.625²

E2 = 23.04 10⁹ Q N / C (1)

now we can solve the two cases presented

a) The total field is

E_total = 50.0 N / C towards + x

since the test charge is positive the electric field E1 points to the right in the direction of the + x axis, so the equation is

E_total = E1 + E₂

E₂ = E_toal - E₁

E₂ = 50.0 -31.10

E2 = 18.9 N /C

With the value of the electric field we can calculate the charge (Q) using equation 1

E₂ = 23.04 10⁹ Q

Q = E₂ / 23.04 10⁹

Q = 18.9 / 23.04 10⁹

Q = 0.820 10⁻⁹ C

the charge on Q is positive

b) E_total = -50.0 N / C

E_total = E₁ + E₂

E₂ = E_total - E₁

E2 = -50.0 - 31.10

E2 = -81.10 N /C

we calculate the charge

Q = E2 / 23.04 10⁹

Q = -81.1 / 23.04 10⁹

Q = -3.52 10⁻⁹ C

for this case the charge is negative