To practice Problem-Solving Strategy 17.2 Electric field calculations Three iden

Question

To practice Problem-Solving Strategy 17.2 Electric field calculations

Three identical particles, q1, q2, and q3, each with charge q = 5.00 C , are placed along a circle of radius r = 2.00 m at angles 90, 180, and 270 degrees from the positive x axis, respectively. What is the resultant electric field at the center of the circle, point O? All angles are measured counterclockwise from the positive x axis.

Problem-Solving Strategy 17.2 Electric field calculations

SET UP

As always, consistent units are essential. With the value of k = 9.00×109 Nm2/C2 , distances must be in meters and charges must be in coulombs. If you are given centimeters or nanocoulombs, don't forget to convert.

Usually, you will use components to compute vector sums. Use proper vector notation; distinguish carefully among scalars, vectors, and components of vectors. Indicate your coordinate axes clearly on your diagram, and be certain that the components are consistent with your choice of axes.

SOLVE

In working out directions of E  vectors, be careful to distinguish between the source point S and the field point P. The field produced by a positive point charge always points in the direction from source point to field point; the opposite is true for a negative point charge.

REFLECT

If your result is a symbolic expression, check to see whether it depends on the variables in the way you expect. If it is numeric, estimate what you expect the result to be and check for consistency with the result of your calculations.

Draw in the three electric field vectors at point O in the diagram. You do not need to worry about absolute magnitudes of the vectors (i.e., whether E 1 should be five or six units long) but you should be sure that the relative magnitudes are accurate (i.e., if E 1 has twice the magnitude of E 2, then you should draw it twice as long).

Explanation / Answer

from the given data,

magnitude electrcif field produced by each charge,
|E1| = |E2| = |E3| = k*q/r^2

= 9*10^9*5*10^-6/2^2

= 1.12*10^4 N/c

direction of E1 is towards -y axis

direction of E2 is towards +x axis

direction of E3 is towards +y axis

here the direction of E1 and E3 are oppisite.

so, net elctric field, Enet = E2

= 1.12*10^4 N/c (towards +x axis) <<<<<<<<<<---------Answer

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