Question

A point charge with a charge q1 = 2.40 μC is held stationary at the origin. A second point charge with a charge q2 = -4.50 μC moves from the point x= 0.140 m , y=0 to the point x= 0.290 m , y= 0.290 m .

Part A

How much work is done by the electric force on q2?

___ J

Answer 1

W = k*q(1)*q(2)*[1/r(1) - 1/r(2)]

k = 9*10^9 Nm^2/C^2

okay, r(1) is the distance from the origin to your starting point
r(1) = .140 m

r(2) is the distance from the origin to your ending point
r(2) = sqrt(.29^2 + .29^2)
r(2) = 0.4101

you have q(1) and q(2) in the question

W = k*q(1)*q(2)*[1/r(1) - 1/r(2)]
W = 9*10^9*2.40*10^-6*(-4.50*10^-6)*(1/.14 - 1/.4101)

W = -0.4572 J

Answer 2

ra=0.170m$r_a=0.170\mathrm{m}$.

rb=(0.250m)2+(0.250m)2−−−−−−−−−−−−−−−−−−−√

rb

EXECUTE: Wa→b=Ua−Ub$W_{a\to b}=U_a-U_b$.

Ua=14πϵ0q1q2ra=(8.99×109N⋅m2/C2)(+3.20×10−6C)(−4.80×10−6C)0.170m.

Ua=−0.8123J$U_a=-0.8123\mathrm{J}$.

Ub=14πϵ0q1q2rb=(8.99×109N⋅m2/C2)(+3.20×10−6C)(−4.80×10−6C)0.3536m

Ub=−0.3906J

Wa→b=Ua−Ub=−0.8123J−(−0.3906J)=−0.422J.