Calculate the magnitude and direction of the electric field intensity

In summary, the conversation discusses the calculation of electric field intensity at a specific point, Y, located 5.00 cm from point X and 8.50 cm from point Z. The superposition principle is used to determine the magnitude and direction of the electric field, with a final result of 3.92 x 10^9 for the magnitude and a direction towards point Y. The conversation ends with gratitude for the help provided.
  • #1
Inquiring_Mike
50
0
Quick Physics Help...

X and Z represent point charges of -2.4 x 10^-3 C and +3.3 x 10^-2 C respectively. Calculate the magnitude and direction of the electric field intensity at point Y, 5.00 cm from X and 8.50 cm from Z (they all live in a line).

Can sumone help me?
 
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  • #2
Apply the superposition principle i.e Electric field at point will be equal to electic field due to X + due to Z.
 
  • #3
So does 4.974 x 10^10 look good?
 
  • #4
you didn't tell about the location of point ,is it left of X or right of Z or in between which will decide mag+dir
 
  • #5
X -----5.0------- Y----------8.5--------Z
 
  • #6
direction will be towards YX and magnitude=9x10^9(2.4x10^-3/5x 10^-2+ 3.3 x 10^-2/ 8.5x10^-2
=3.92 x 10^9
 
  • #7
Ah... It all makes sense now...
THanks for the help!
 

FAQ: Calculate the magnitude and direction of the electric field intensity

What is the electric field intensity?

The electric field intensity is a physical quantity that describes the strength and direction of an electric field at a particular point in space. It is a vector quantity, meaning it has both magnitude and direction.

How is the electric field intensity calculated?

The electric field intensity is calculated using the formula E = kQ/r^2, where E is the electric field intensity, k is the Coulomb's constant (9 x 10^9 Nm^2/C^2), Q is the charge creating the electric field, and r is the distance from the charge to the point where the electric field is being measured.

What are the units of electric field intensity?

The units of electric field intensity are newtons per coulomb (N/C) in SI units. In other systems of units, such as the CGS system, the units are dynes per statcoulomb (dyn/statC).

How does the direction of the electric field intensity relate to the direction of the electric force?

The direction of the electric field intensity is always in the direction that a positive test charge would experience an electric force. This means that the electric field and electric force vectors are parallel to each other.

How does the electric field intensity change with distance from a point charge?

The electric field intensity decreases with distance from a point charge according to an inverse square relationship. This means that as the distance from the charge increases, the electric field intensity decreases by a factor of the square of the distance.

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