Question
Two carbon discs, 1.0 g each are 1.0 cm apart have equal and opposite charges. If the force of attraction between them is 10–5 N, then the ratio of excess electrons to the total atoms on the negatively charged disc is (NA = 6 × 1023, e = 1.6 × 10−19 C)
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Solution
Number of atoms in the disc $\frac{1}{12} \times 6 \times 10^{23} = 5 \times 10^{22}$
Now, $F = K.\frac{q_{1}q_{2}}{r^{2}} \Rightarrow 10^{- 5} = 9 \times 10^{9} \times \frac{q^{2}}{\left( 10^{- 2} \right)^{2}} \Rightarrow q = \frac{10^{- 10}}{3}C$
∴ Number of excess electron on negatively charged disc = $\frac{10^{- 10}/3}{1.6 \times 10^{- 19}} = \frac{10^{9}}{4.8}$
Hence, $\frac{\text{Number of excess electron}}{\text{Number of atoms}} = \frac{10^{9}/4.8}{5 \times 10^{22}} = \frac{10^{- 14}}{2.4}$
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