Three charges are fixed to an x , y coordinatesystem. A charge of +18 µC is on t

Question

Three charges are fixed to an x, y coordinatesystem. A charge of +18 µC is on the y axis aty = +3.0 m. A charge of -12 µC is atthe origin. Lastly, a charge of +41 µC is onthe x axis at x = +3.0 m. Determine the magnitudeand direction of the net electrostatic force on the charge atx = +3.0 m. Specify the direction relative to the-x axis. magnitude 1 N direction 2° 3---Select---abovebelow the -x axis I don't have any idea where to start. There seems like so manysteps that I'm getting confused. Please someone walk me throughthis whole problem. I'm really trying to understand. (there will bea problem similar to this on our exam worth 25% of the examgrade) Thank you :) magnitude 1 N direction 2° 3---Select---abovebelow the -x axis I don't have any idea where to start. There seems like so manysteps that I'm getting confused. Please someone walk me throughthis whole problem. I'm really trying to understand. (there will bea problem similar to this on our exam worth 25% of the examgrade) Thank you :) magnitude 1 N direction 2° 3---Select---abovebelow the -x axis

Explanation / Answer

Well. so this kind of question. You should do like this.     Figure out the force (vector) that act on thecharge. It is best to draw a gragh, draw on it the direction, magnitude,etc....... The direction should be obtain from the relative position of thecharges and it sign. Draw it all for easier doing     The second: calc the magnitude of the force. The magitude is obtained from the equation F=kQq/r^2     Now assume that you have know all the directionand magitude of every invidual force that act on that charge. There are 2 ways to do thing from here:     First: Try to figure out the x-component andy-component of every forces act on that charge. When you done, just add the x-components together, the y-componenttogether, and the angle should be tan=Fy/Fx and the magnitude is sqrt(Fy^2+Fx^2)     Second: Using vector calculation.

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