Electrostatic field ( Gauss Law )

AI Thread Summary
To determine the electric intensity at the point (-1, 5, -3) due to an infinitely long line charge of 50 pC/m located at (2, y, 1), the relevant equation is E = (ρ / (2πε₀r)). The user initially calculated the distance vector incorrectly by focusing on the point P to Q instead of the perpendicular distance from point P to the line charge. Clarification was provided that visualizing the setup would help understand the correct distance to use in the formula. The final answer for the electric intensity is -0.18(ax0.6 + az0.8)(V/m).
Saru
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Homework Statement


Assuming that an infinitely long line charge of 50(pC/m) parallel to the y-axis at x=2(m) and z=1(m), determine the electric intensity at the point (-1, 5, -3).

The answer given : -0.18(ax0.6 + az0.8)(V/m)

Homework Equations


Electric field intensity due to an infinite straight line charge of uniform density :
E= ar("rho" / (2"pi""epsolon 0"r) ( V/m)

The Attempt at a Solution



I let point P = -ax + ay5 - az3
Point on line perpendicular to point P, Q = ax2 + az

Vector Q to P = -ax3 + ay5 - az4

I found out that if i ignore or make coefficient of ay to zero and apply the above formula i'll get the answer..

but i do not understand why..please help
 
Last edited:
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Hi Saru,

Draw a picture and you will see the reason immediately. You are not looking for the distance from P to Q, but from P to the line of charge.
 
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