Answer to Question #311440 in Electric Circuits for [pip;

Question #311440

A – 41 nC charge is placed in between two-point charges, 32 nC and 24 nC. The negative

charge moves from point a to b. What is the change in potential energy?


1
Expert's answer
2022-03-16T09:42:52-0400

Solution;


Find the potential at point a;

Due to 32nC;

U32,41=kq1q2rU_{32,-41}=\frac{kq_1q_2}{r}

U=8.99×109×32×41×10180.1=1.179×104JU=\frac{8.99×10^9×32×-41×10^{-18}}{0.1}=1.179×10^{-4}J

Due to the 24nC;

U=8.99×109×41×24×10180.05=1.769×104JU=\frac{8.99×10^9×-41×24×10^{-18}}{0.05}=1.769×10^{-4}J

Total potential energy at point a is;

Ua=(1.179+1.769)×104=2.948×104JU_a=(1.179+1.769)×10^{-4}=2.948×10^{-4}J

After moving to b;

Distance between the 32nC is;

r=102+72=0.122mr=\sqrt{10^2+7^2}=0.122m

Potential energy becomes;

U=8.99×109×41×32×10180.122=0.9667×105JU=\frac{8.99×10^9×41×32×10^{-18}}{0.122}=0.9667×10^{-5}J

Distance between the 25nC becomes;

r=52+72=0.0862r=\sqrt{5^2+7^2}=0.0862

Potential energy becomes;

U=8.99×24×41×10180.0862=1.0262×104JU=\frac{8.99×24×41×10^{-18}}{0.0862}=1.0262×10^{-4}J

Total potential energy at point b is;

Ub=(1.0262+0.9667)×104=1.9929×104JU_b=(1.0262+0.9667)×10^{-4}=1.9929×10^{-4}JJ

The difference in potential difference is;

ΔU=UaUb\Delta U=U_a-U_b

ΔU=2.9841.9924=0.9911×104J\Delta U=2.984-1.9924=0.9911×10^{-4}J


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment