I need to know how to work these problems: 1. Two charges are located on the x a
ID: 1706386 • Letter: I
Question
I need to know how to work these problems:
1. Two charges are located on the x axis: q1= +5.0C at x1=+4.0cm, and q2 = +5.0C at x2 = -4.0cm. Two other charges are located on the y axis: q3= +1.0C at y3= +5.0cm, and q4= -5.0C at y4 = +7.0cm. Find the net electric field (magnitude and direction) at the origin.
magnitude ____________N/C
direction____________ degrees (measured counter clockwise from the x axis)
2. At a distance r1 from a point charge, the magnitude of the electric field created by the charge is 238 N/C. At a distance r2 from the charge, the field has a magnitude of 131 N/C. Find the ratio of the distances.
r2/r1=____________
Thank you in advance and it would be very much appreciated if you can explain the process please. thanks!!
Explanation / Answer
(1) Electric field due to a point charge, E = k q / r^2 Electric fields at the origin due to q1 and q2 are equal and opposite. So, they cancel each other. Electric field due to q3, E3 = k q3 / r^2 = 9 x 10^9 * 1 x 10^-6 / ( 0.05 )^2 = 3.6 x 10^6 N/C ( towards -ve y-axis ) Electric fields due to q4, E4 = k q4 / r^2 = 9 x 10^9 * 5 x 10^-6 / ( 0.07 )^2 = 9.2 x 10^6 N/C ( towards +ve y-axis ) Net electric field at origin, E = E4 - E3 = 5.6 x 10^6 N/C ( towards + ve y-axis ) Magnitude = 5.6 x 10^6 N/C Direction: 90 degrees ( counterclockwise from x-axis ) (2) Electric field is inversely proportional to the square of distance from the given charge. E ( 1 / r^2 ) ( r2 / r1 ) = ( E1 / E2 ) = ( 238 / 131 ) = 1.35 Ratio of distances = 1.35 : 1Related Questions
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