physics help A +5.5×10-7 C charge is located at the origin.?
A +5.5×10-7 C charge is located at the origin. A -9.4×10-6 C charge is located at +0.43 m on the x-axis. A +6.2×10-6 C charge is located at +0.65 m on the y-axis. Find the magnitude only (no direction needed) of the net force acting on the charge at the origin.
Three charges, A, B, and C, are placed on a line. Charge A, with a magnitude of -9.6x10-6 C is on the left. Charge B, with a magnitude of +7.3x10-5 C is 0.46 m to the right of A. Charge C, with a magnitude of +3.8x10-5 C is 0.69 m to the right of B. Find the net force acting on C.
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Hello Kevin Criswell, let me answer only one at a time. Okay?
Due to -ve charge attractive force along +ve x direction
whose magnitude = 9*10^9 * (5.5*9.4)*10^-13 / (0.43)^2 N
And due to +ve charge a repulsive force in -ve y direction with magnitude
9*10^9 * (5.5*6.2)*10^-13 / (0.65)^2 N
As these two forces are perpendicular to each other their resultant
R = ./ (F1)^2 + (F2)^2
Any way 9*10^9 * 5.5 * 10^-13 are found to be common
Only this is to simplified
./ (9.4)^2 / (0.43)^4 + (6.2)^2 / (0.65)^4
So it comes to be 52.9
Now use this along with 9*10^9 * 5.5 * 10^-13 and you will get the required of the net force on the charge on the origin.