Four equal charges of +3.5×10 -6 C are placed on the corners of one face of a cu
ID: 1404647 • Letter: F
Question
Four equal charges of +3.5×10-6 C are placed on the corners of one face of a cube of edge length 5.0 cm. A charge of -3.5×10-6 C is placed at the center of the cube. What is the magnitude of the force on the charge at the center of the cube?Four equal charges of +3.5×10-6 C are placed on the corners of one face of a cube of edge length 5.0 cm. A charge of -3.5×10-6 C is placed at the center of the cube. What is the magnitude of the force on the charge at the center of the cube?
Four equal charges of +3.5×10-6 C are placed on the corners of one face of a cube of edge length 5.0 cm. A charge of -3.5×10-6 C is placed at the center of the cube. What is the magnitude of the force on the charge at the center of the cube?
Explanation / Answer
Lenth of body diagonal of cube is sqrt(3)*a
Where a is edge length.
So, distance of each charge form centre charge = sqrt(3)*a/2 = sqrt(3)*5/2 = 4.33 cm = 0.0433 m
If F1,F2,F3 and F4 are forces fue to each charge,
we should know that one Componet of each force is cancelled out by others. So magnitude of net force is not equal to sum of magnitude of all Forces.
Instead It will be sum of magnitude of all forces * sin 45
Since all forces are same and sin45 is 1/sqrt(2).
Magnitude of net force = 4*F*1/sqrt(2)
= 2*sqrt(2)*F
Here F =k*Q1*Q2/R^2
= (9*10^9)(3.5*10^-6)*(-3.5*10^-6) / (0.0433)^2
= -58.8 N
So magnitude of net force = 2*sqrt(2)*F = 2*sqrt(2)*58.8 = 166 N
Answer: 166 N
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