Question

A point charge with charge q1 = 2.10 μC is held stationary at the origin. A second point charge with charge q2 = -4.70 μC moves from the point ( 0.105 m , 0) to the point ( 0.290 m , 0.245 m ). How much work W is done by the electric force on the moving point charge?

Answer 1

Answer:

the work W required is 0.611 J

Explanation:

at the beginning, the distance d between the charges at coordinates (x₁=0.105 m,y₁=0) for charge 1 and  (x₀=0,y₀=0) for charge 2 is:

d₁ = √[(x₁-x₀)² +(y₁-y₀)²] = √[(0.105 m-0)² +(0-0)²] = 0.105 m

when the charge 1 moves to the  (x₂=0.290 m , y₂=0.245 m ), the new distance to charge 2 will be:

d₂ = √[(x₂-x₁)² +(y₂-y₁)²] =  √[(0.290m-0)² +(0.245m-0)²] = 0.3796 m

knowing the initial distance and the final distance, we can now calculate the work W done by the electric field

W = k*q1*q2*(1/d₂-1/d₁)

where k= Coulomb's constant= 8.988×109 N⋅m²/C²

replacing values

W = k*q1*q2*(1/d₂-1/d₁)= (8.988*10⁹ N⋅m²/C²)*(2.10*10⁻⁶ C )*(-4.70*10⁻⁶ C)*[1/0.3796 m- 1/0.105 m ] = 0.611 N*m = 0.611 J

Answer 2

Answer:

0.613 J

Explanation:

charge q1 = 2.10 μC = $2.1 x 10^{-6} C$

charge q2 = -4.70 μC = $-4.7 x 10^{-6} C$

location of q1 = (0,0)

initial location of q2 = (0.105m , 0)

final location of q2 =  (0.290m , 0.245m )

find the work done by the electric forces on the moving point charge

work done = kq1q2($\frac{1}{R2} - \frac{1}{R1}$)

where

• R1 = $\sqrt{0.105^{2} + 0^{2} }$ = 0.105

• R2 = $\sqrt{0.290^{2} + 0.245^{2} }$ = 0.38
• k = $\frac{1}{4πE₀}$ = $9 x 10^{9} N.m^{2}/C^{2}$

work done =  $9 x 10^{9} x 2.1 x 10^{-6} x -4.7 x 10^{-6} x (\frac{1}{0.38} - \frac{1}{0.105})$

work done = -88 x 10^{-3} x -6.9 = 0.613 J