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

0.613 J

Step-by-step 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(
`(1)/(R2) - (1)/(R1)`)

where

• R1 =
  `\sqrt{0.105^(2) + 0^(2) }` = 0.105

• R2 =
  `\sqrt{0.290^(2) + 0.245^(2) }` = 0.38
• k =
  `(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 ((1)/(0.38) - (1)/(0.105))`

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

Answer 2

the work W required is 0.611 J

Step-by-step 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