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And what is the value of ##x## from that equation?
The discussion centers on determining the neutral point between two electric charges, specifically using the electric field vector equation. Participants debate the validity of writing the equation as E1 vector + E2 vector = 0 versus using magnitudes. The correct approach involves expressing the electric field at an arbitrary point P using unit vector notation and the formula $$\mathbf E(\mathbf r)=\frac{1}{4\pi\epsilon_0}\sum_{i=1}^2\frac{q_i(\mathbf r-\mathbf r'_i)}{|\mathbf r-\mathbf r'_i|^3}$$. The neutral point must lie along the line connecting the two charges, which simplifies the calculations.
PREREQUISITESStudents of physics, particularly those studying electromagnetism, educators explaining electric field concepts, and anyone involved in solving problems related to electric charges and fields.
I get x as ## \frac {√2} {3} ## @kurumantellmesomething said:Yes $$ \frac {kq} {x²}= \frac {k4q} {(√2-x)²} $$
Further I can multiply this with the unit vector and find the vevtor joining r2 and r3 as well as r3 and r1kuruman said:And what is the value of ##x## from that equation?
And from there its easy to find out the position vector r3tellmesomething said:Further I can multiply this with the unit vector and find the vevtor joining r2 and r3 as well as r3 and r1
Go ahead and do it.tellmesomething said:And from there its easy to find out the position vector r3
Done! Got the answer.kuruman said:Go ahead and do it.