Four identical charges each of charge are placed at the corners of a square. Then at the centre of the square the resultant electric intensity E and the net electric potential V are
E≠0,V=0
E=0, V=0
E=0,V≠0
E≠0,V≠0
A
E=0, V=0
B
E≠0,V≠0
C
E≠0,V=0
D
E=0,V≠0
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Solution
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→ electic field
by the diagonality opposits changes canal each other
∴E=0
voltage:
we know V=KVe
( E is vector V is a scalar )
V=√2KVa
total V=4√2KVa
∴E=0,V≠0
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