What is the Net Electric Field Vector at a Point Due to Two Charges?

AI Thread Summary
The discussion focuses on calculating the net electric field vector at point A, located at <0, 0, 0>m, due to two equal charges of +6 μC positioned at <−1, 0, 0>m and <5, 0, 0>m. The electric field equations used are E=KQ/|r|^2 * |r|, with calculations for the electric fields E1 and E2 from each charge. E1 is calculated as <54,0,0>, while E2 requires recalculation and is initially found to be <.93312,0,0>. The net electric field vector is determined by summing E1 and E2, resulting in a final vector of <54.93312,0,0>. The discussion emphasizes the importance of accurate calculations for E2.
perfection256
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Homework Statement



Two equal charges +6 μC are placed, one at <−1, 0, 0>m and the other at <5, 0, 0> m. What is the net electric
field vector at the location A which is at <0, 0, 0>m? Show your work.

Homework Equations



E=KQ/|r|^2 * |r|

The Attempt at a Solution



r1=<0,0,0>-<-1,0,0>=<1,0,0>

E1= k*(6x10^-9)/1 * <1,0,0>= <54,0,0>

r2= <0,0,0>-<5,0,0>= <-5,0,0>

E2= k*(6x10^-9)/(5^2) * <10.8/25,0,0>= <.93312,0,0>

E1+E2=ENet

Enet= <54,0,0>+<.93312,0,0>=

<54.93312,0,0>
 
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perfection256 said:

Homework Equations



E=KQ/|r|^2 * |r|
This should be:

E=KQ/|r|^2 * \hat r

The Attempt at a Solution



r1=<0,0,0>-<-1,0,0>=<1,0,0>

E1= k*(6x10^-9)/1 * <1,0,0>= <54,0,0>
OK.

r2= <0,0,0>-<5,0,0>= <-5,0,0>

E2= k*(6x10^-9)/(5^2) * <10.8/25,0,0>= <.93312,0,0>
Recalculate E2.
 
Thanks Doc Al!
 
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