Solution;
Find the potential at point a;
Due to 32nC;
U32,−41=rkq1q2
U=0.18.99×109×32×−41×10−18=1.179×10−4J
Due to the 24nC;
U=0.058.99×109×−41×24×10−18=1.769×10−4J
Total potential energy at point a is;
Ua=(1.179+1.769)×10−4=2.948×10−4J
After moving to b;
Distance between the 32nC is;
r=102+72=0.122m
Potential energy becomes;
U=0.1228.99×109×41×32×10−18=0.9667×10−5J
Distance between the 25nC becomes;
r=52+72=0.0862
Potential energy becomes;
U=0.08628.99×24×41×10−18=1.0262×10−4J
Total potential energy at point b is;
Ub=(1.0262+0.9667)×10−4=1.9929×10−4JJ
The difference in potential difference is;
ΔU=Ua−Ub
ΔU=2.984−1.9924=0.9911×10−4J
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