A thin conducting rod AB is introduced in between the two point charges + $q_{1}$ & - $q_{2}$ as shown in figure. For this situation mark the correct statement (s)

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The total force experienced by $q_{1}$ is vector sum of electric force experienced by. $q_{1}$ due to $q_{2}$ and due· to induced charges on rod .

The end A will become negatively charged

The total magnitude of force acting on +q₁ will be greater than as compared to the case without rod

The total force acting on - $q_{2}$ will be greater than as compared to the case without rod

answer is A, B, C, D.

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Detailed Solution

Due to induction effect the situation is shown clearly in figure.

Due to $+ q_{1}$ let induced charge is $- q_{1}^{'}$ at end A and $+ q_{1}^{'}$ at end B, while due to - $q_{2}$ , induced charges are $- q_{2}^{'} & + q_{2}^{'}$ at ends A & B respectively.

Thus the end A acquires negative charge and B acquires positive charge.

Electric force experienced by q₁ or - q₂ has to be computed by using principle of superposition. Force on $+ q_{1}$ due to - $q_{2}$ towards right. Due to rod also towards right. Hence total force experienced by + $q_{1}$ in present situation is greater than as compared to the case without rod.

Same is the situation for - $q_{2}$ .

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