Question #298325

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.


1
Expert's answer
2022-02-17T04:34:39-0500

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 Ei=kqiri2ri^\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 r1^=i^ and r2^=i^\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 r1^=r2^=i^\hat{r'_1}=\hat{r'_2}=\hat{i}. Using that information we find the total electric field as ET=E1+E2:


  • for x=1m:


(9×109Nm2/C2)(q1(1m)2q2(1m)2)i^=(10.8N/C)i^(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:

(9×109Nm2/C2)(q1(3m)2+q2(1m)2)i^=(8N/C)i^(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:

q1q2=1.2×109Cq1+9q2=8×109Cq_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.


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