Question

A positive point charge q1 at the origin and a negative charge q2 at x=2m. The resultant electric field is 10.8 i N/C at x=1m and -8 i N/C at x=3m. Find q1 and q2.

Answer 1

We need to establish the two equations to determine the total electric field at x=1m and x=3m. The electric field is equal to $\vec{E_i}=k\dfrac{q_i}{r_i^2} \hat{r_i}$. We know that q₁>0 and q₂<0, k=9X10⁹ Nm²/C², and for the first case at x=1 m, r₁=r₂=1m while $\hat{r_1}=\hat{i} \text{ and } \hat{r_2}=-\hat{i}$. On the second case at x=3m we have r'₁=3m, r'₂=1m, and $\hat{r'_1}=\hat{r'_2}=\hat{i}$. Using that information we find the total electric field as E_T=E₁+E₂:

• for x=1m:

$(9\times10^9 Nm^2/C^2)\Big( \frac{q_1}{(1\,m)^2} -\frac{q_2}{(1\,m)^2} \Big)\hat{_i}=(10.8\,N/C) \hat{_i}$

• for x=3m:

$(9\times10^9 Nm^2/C^2)\Big( \frac{q_1}{(3\,m)^2} +\frac{q_2}{(1\,m)^2} \Big)\hat{_i}=(-8\,N/C) \hat{_i}$

The system that has to be solved is:

$q_1-q_2=1.2\times10^{-9}\,C \\ q_1+9q_2=-8\times10^{-9}\,C$

After solving, in conclusion, q₁ = 2.8 X 10^-10 C and q₂= -9.2 X 10^-10 C.