Find the potential energy of a system of four identical particles placed at the vertices of a square of side l. Also obtain the potential at the centre of the square.
Answers
The system can be made by bringing each charge from reference (∞) and
placing them on the vertices of the square
Then net work done in arranging the system = Potentital energy of teh system
For bringing 1st charge : W
1
=0
For bringing 2nd charge : W
2
l
2
kqq
(assuming each charge = q and side = l)
For bringing 3rd charge : W
3
=
l
2
kqq
+
(
2
l)
2
kqq
For bringing 4th charge: W
4
=
l
2
kqq
+
l
2
kqq
+
(
2
l)
2
kqq
Hence total work done = W
1
+W
2
+W
3
+W
4
=
l
2
5kq
2
i.e. potential energy of system
The potential at centra = 4×
(
2
1
)
2
kq
=
l
2
8kq
Explanation:
The system can be made by bringing each charge from reference (∞) and
placing them on the vertices of the square
Then net work done in arranging the system = Potentital energy of teh system
For bringing 1st charge : W1 = 0
For bringing 2nd charge : W2 = kqq/l^2
(assuming each charge = q and side = l)
For bringing 3rd charge : W3 = kqq/l^2 + kqq/(✓2l)^2
For bringing 4th charge: W4 = kqq/l^2 + kqq/l^2 + kqq/(✓2l)^2Hence total work done = W1 + W2 + W3 + W4 = 5kqq/l^2 i.e. potential energy of system
The potential at centre = 4× kq/(l/✓2)^2 = 8kq/l^2