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:
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