A neutral ammonia $\left(\mathrm{NH}_3\right)$ in its vapour state has an electric dipole moment of…

A neutral ammonia $\left(\mathrm{NH}_3\right)$ in its vapour state has an electric dipole moment of magnitude $5 \times 10^{-30} \mathrm{C}$-m. How far apart are the molecules centres of positive and negative charge.
  1. $4.125 \times 10^{-12} \mathrm{~m}$
  2. $3.125 \times 10^{-12} \mathrm{~m}$
  3. $3.125 \times 10^{-6} \mathrm{~m}$
  4. $4.125 \times 10^{-6} \mathrm{~m}$

Solution

Dipole moment, $\mathrm{p}=5 \times 10^{-30} \mathrm{C}-\mathrm{m}$
The charge on neutral $\mathrm{NH}_3$ is $\mathrm{q}=7+1 \times 3=10 \mathrm{e}$ $\begin{aligned} & \therefore \quad \mathrm{p}=\mathrm{q}(1) \\ & \Rightarrow 5 \times 10^{-30}=10 \times 1.6 \times 10^{-19} \times 1 \\ & \therefore \quad 1=\frac{5 \times 10^{-30}}{16 \times 10^{-19}}=3.125 \times 10^{-12} \mathrm{~m} \end{aligned}$

Asked in: AP EAMCET 2024 (22 May Shift 1)

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