Daniel B. answered • 02/23/22
A retired computer professional to teach math, physics
Let
q = 10 μC be each of the charges,
r = 3 m be the distance the charges are from the origin,
k = 9×109 kg⋅m³/s²C² be the electrostatic constant.
1.
The question does not define "assemble".
Potential energy is defined as the work required to bring the charges
from infinity. However, the question does not mention potential energy.
Nevertheless, I will explain the answer under the assumption that
"assemble" means bringing from infinity.
If "assemble" means something else, hopefully you may be able to modify my
solution to solve it under the correct meaning of "assemble".
Imagine that both charges start at infinity, and we move them in a symmetrical
way to their positions at (-r, 0) and (r, 0).
When they are at some intermediate positions (-x, 0), (x, 0)
then they repulse each other with the force
F = kq²/(2x)²
Therefore the work required is the definite integral from infinity to r
W = ∫Fdx =
∫kq²/(2x)² dx =
kq²/4 ∫1/x² dx =
kq²/4 (-1/x)
For x=∞ the above expression evaluates to 0.
Therefore the total work is
W = kq²/4r
Substituting actual numbers
W = 9×109 ×(10×10-6)²/(4×3) = 0.075 J
2.
The distance, d, from either of the two charges to the point (0,3) is
d = 3√2
The magnitude of the electric field generated by either of the two charges
at the point (0,3) is defined as
E0 = kq/d²
The electric field from both charges combined is the vector sum of the individual fields.
The two vectors are perpendicular to each other and of the same magnitude E0.
Therefore their sum will have the magnitude
E = E0√2 = kq√2/d²
Substituting actual numbers
E = 9×109 ×10×10-6√2/(3√2)² = 7071 N/C
3.
The force on charge Q=1μC at (0,3) is
F = EQ
Substituting actual numbers
F = 7071 × 10-6 = 0.007071 N
