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Three identical charges $(q=-5.0 \mu \mathrm{C})$ lie along a circle of radius 2.0 $\mathrm{m}$ at angles of $30^{\circ}, 150^{\circ},$ and $270^{\circ},$ as shown in Figure $\mathrm{P} 15.99$ (page 524$) .$ What is the resultant electric field at the center of the circle?

The net field at the center of the circle is 0

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Rutgers, The State University of New Jersey

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McMaster University

for this problem on the topic of electric forces and fields were given three identical charges, which are lying along a circle of a given radius. We know the angle at which the charges lie as shown in the figure, and we want to find the resultant electric field at the center of the circle. Now, from symmetry of charge distribution, we can recognize that the resultant electric field at the center E. R. Is equal to zero. And if you want to show that this is true, then we can take the resultant electric field, which we know should be electric fields to charge one e one last the electric field to charge to eat, too, plus the electric field due to charge the E three, so we'll calculate them in components. E X is equal to e one x minus. E two x e three has no horizontal or X component, and this is okay. Mm times the magnitude of charge Q. Over R Squared times the call sign of 30 degrees minus K E E. Times Q over R Squared times, the co sign of 30 degrees, and this is equal to zero Now. If we calculate the white components of the resultant electric field. This is E one y plus e two y minus E three e three only has a vertical while component. So there's is K E E Times Q. OF, UH r squared times The sign of 30 degrees plus k e times Q over R squared times the sine of 30 degrees minus. Okay, I e times Q over our squared. And if we calculate this, we can see that this is equal to zero. So we have therefore that the magnitude of the results and electric field E R, which is the square root of the sum of squares with the X and Y components he x squared. Let's see y squared is equal 20

University of Kwazulu-Natal