anonymous
  • anonymous
wo charges, Q1 = 4.00 μC and Q2 = 6.80 μC, are located at points (0, -2.65 cm) and (0, +2.65 cm), as shown in the figure. What is the magnitude of the electric field at point P, located at (6.15 cm, 0), due to Q1 alone?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
|dw:1349471622020:dw|
anonymous
  • anonymous
@demitris could u help me with this one last question,and thanks for the others u did.got it right
anonymous
  • anonymous
What is the x-component of the total electric field at P? What is the y-component of the total electric field at P? What is the magnitude of the total electric field at P?

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anonymous
  • anonymous
Did you get your answer?
anonymous
  • anonymous
NOT YET :(
anonymous
  • anonymous
GOT ONLY NUMBER 1
anonymous
  • anonymous
|dw:1349477252753:dw| E1 is electric field due to q1. \[E_1=\frac{1}{4\pi\epsilon_0 }\frac{q_1}{r^2} \] making angle \[\theta=\arctan{\frac{y}{x}}\]with horizontal.
anonymous
  • anonymous
Okay, now here's the strategy: 1) Find the magnitude of E1 first (you already did it in no 1) 2) Find the magnitude of E2 (no useto do this, since both q1 and q2 has the same charge) 3) Find x component of E1 and E2 (actually you just need to find x component of E1. E2 has the same x component). 4) Find y component of E1 and E2 (the total y component of E1 and E2 cancels each other. Since they have same magnitude but opposite direction)
anonymous
  • anonymous
@imron07 not quite right. q1 and q2 are different.
anonymous
  • anonymous
Haha, thanks @Algebraic! . I thought they have common charge.
anonymous
  • anonymous
|dw:1349681562544:dw|
anonymous
  • anonymous
r^2=(X2-X1)^2-(Y2-Y1)^2 distance formula

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