Question

A point charge q1=+2.40μC is held stationary at the origin. A second point charge q2=−4.30μC moves from the point x=0.125m , y=0, to the point x=0.275m , y=0.275m.
Part A

What is the change in potential energy of the pair of charges?

Express your answer in joules to three significant figures.

Part B

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

Express your answer in joules to three significant figures.

Answer 1

Answer:

Explanation:

Potential energy of a pair of charges

= k q₁q₂ /r₁₂

At r₁₂ = .125 m

= - 9 x 10⁹ x 2.4 x 4.3 x 10⁻¹² / .125

= - 743 x 10⁻³J

At r₁₂ = √2 x .275 m

- 9 x 10⁹ x 2.4 x 4.3 x 10⁻¹² / √2 x .275

= - 238.85  x 10⁻³J

= 239 x 10⁻³ J

Change = - 238.85  x 10⁻³ +743 x 10⁻³

= 504.15 x 10⁻³ J

B )

Work done by electric force will be same but in negative sign

= - 504.15 x 10⁻³ J

= - 504 x 10⁻³ J