Anonymous
Anonymous asked in Science & MathematicsPhysics · 8 months ago

# A point charge with charge q1 = 3.90 μC is held stationary at the origin. ?

A second point charge with charge q2 = -4.50 μC moves from the point ( 0.170 m , 0) to the point ( 0.265 m , 0.285 m ). How much work W is done by the electric force on the moving point charge? Relevance
• The voltage at (0,0.17) = kq/r = k*3.9e-6/0.17 = 206,185V

The voltage at (0.265,0.285) =

k*3.9e-6/√(0.265²+0.285²) = 90,068V

ΔV = 90,068 - 206,185 = -116,117V

W= q*ΔV = -4.5e-6 * -116,117 = +0.523J which makes sense because the negative charge will have to be forced away from the positive charge which is positive work.

• Anonymous
8 months ago

d = 0.17 m

F = 9.0*10^9*3.9*4.5*10^-12*10^4/17^2 = 5.47 N

d' = √0.265^2+0.285^2 = 0.39 m

F' = 9.0*10^9*3.9*4.5*10^-12*10^4/39^2 = 1.04 N

Δd = √(0.265-1.7)^2+0.285^2 = 0.31 m

work W = (F-F')/3*Δd = (5.47-1.04)/3*0.31 = 0.46 joule

• PE = kQq / d

so

work = ΔPE = 8.99e9N·m²/C² * 3.90e-6C * -4.50e-6C * (1/0.170m - 1/√(0.265² + 0.285²)m)

work = -0.523 J