A charge of 4 C is on the y-axis at 4cm (0,4) and a second charge of -4 C is on

Question

A charge of 4 C is on the y-axis at 4cm (0,4) and a second charge of -4 C is on the y axis at -4cm (0,-4)

Find the force on a charge of 1C on the x-axis at x= 7cm. The value of the Coulomb constant is

8.98755 x 109 Nm2/C2

   I started by making a graph on an XY plane, then calculated the distance from p1 (0,4) to p3(7,0) to be 8.062257748 cm (the home work program is extremely picky about rounding). I used that to find the angle between p3 and p1 to be 22.74. Using the formula f=k(q1q2)/r2 I figured out the magnitude of force, split it into componant vectors I and J, and added them to the p2 using symmetry. All of my answers that I input are wrong though. Please help

Explanation / Answer

Let q1 = 4 C and is at (0, 4cm) and q2 = -4 C and is at (0, -4cm) Let the third charge q3 = 1 C is at 7cm on positive X-axis. Now let the force of repulsion between q1 and q3 is F1 is given by    F1 = kq1q3/r1^2 here k = 8.98755*10^9 Nm^2/C^2 ; r1 = sqrt(4^2 + 7^2) = 8.1 cm = 0.081 m F1 = 5.49 * 10^-6 N The force of attraction between q2 and q3 is F2 and is given by  F2 = kq1q3/r2^2 here k = 8.98755*10^9 Nm^2/C^2 ; r1 = sqrt(4^2 + 7^2) = 8.1 cm = 0.081 m F2 = 5.49 * 10^-6 N The angle between F1 and F2 is 90 degrees So the resultant force, F = sqrt (F1^2 + F2^2) = 7.8 * 10^-6 N here k = 8.98755*10^9 Nm^2/C^2 ; r1 = sqrt(4^2 + 7^2) = 8.1 cm = 0.081 m F2 = 5.49 * 10^-6 N The angle between F1 and F2 is 90 degrees So the resultant force, F = sqrt (F1^2 + F2^2) = 7.8 * 10^-6 N

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