A) 8e
B) 6e
C) 3e
D) e
Correct Answer: C
Solution :
For equilibrium weight of particle = electrostatic force \[mg=qE\] \[mg=q\left( \frac{V}{d} \right)\] Putting the given values we get \[1.96\times {{10}^{-15}}\times 9.8=q\left( \frac{800}{0.02} \right)\] \[q=\frac{1.96\times {{10}^{-15}}\times 9.8\times 0.02}{800\times 1.6\times {{10}^{-19}}}e\]\[=3e\]You need to login to perform this action.
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