1 If a test charge ap = - 12X10-9 %
is placed at the centre of line
joining the two charges.
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Answer:
Net repulsive force on any Q charge particle due to another charge particle =
r
2
K(Q)
2
net attractive force on any charge particle =
(r/2)
2
KQq
net force will be zero so
r
2
K(Q)
2
+
(r/2)
2
KQq
=0
q=−Q/4
Since q is placed symmetrically in between both charges, there won't be any unbalanced force on q
Explanation:
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