3 identical charges of 10mc each are placed at the 3 vertices of an equilateral triangle of sides 1m.the work to be done in moving oneof them to the mid points of the line joining the other two is
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Three particles are situated at A,B and C having charge Q=10μC each.
The distance between the two charges r=10cm=0.1 m
Potential energy of the charges at A and B, E
AB
=
4πϵ
o
1
r
Q
2
∴ E
AB
=
0.1
9×10
9
×(10×10
−6
)
2
=9J
Similarly, E
BC
=E
AC
=9J
Total potential energy of the system, E=E
AB
+E
BC
+E
AC
=27J
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