A) 1 : 2
B) 1 : 3
C) 2 : 1
D) 3 : 1
Correct Answer: B
Solution :
[b] \[{{T}_{AB}}=\frac{kq\cdot 2q}{{{d}^{2}}}+\frac{kq\cdot 4q}{{{\left( 2d \right)}^{2}}}=\frac{3k{{q}^{2}}}{d}\] \[{{T}_{BC}}=\frac{k\cdot 4q\cdot 2q}{{{d}^{2}}}+\frac{kq\cdot 4q}{{{\left( 2d \right)}^{2}}}=\frac{9k{{q}^{2}}}{d}\]You need to login to perform this action.
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