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

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SUMMARY

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, is calculated using the formula E=KQ/|r|^2 * |r|. The electric field contributions from each charge are E1=<54,0,0> and E2=<.93312,0,0>. The resultant net electric field vector is ENet=<54.93312,0,0>.

PREREQUISITES
  • Understanding of Coulomb's Law and electric fields
  • Familiarity with vector addition in three-dimensional space
  • Knowledge of the constant K (Coulomb's constant)
  • Basic calculus for understanding vector magnitudes and directions
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  • Review the derivation of Coulomb's Law and its applications
  • Study vector addition techniques in physics
  • Learn about electric field lines and their significance
  • Explore the concept of superposition in electric fields
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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 * [itex]\hat r[/itex]

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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