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Step-by-Step Solution
Step 1: Recall the formula for electric potential due to a short dipole
The electric potential at a point making an angle $ \theta $ with the dipole axis, at a distance $ r $ from the center of a short electric dipole of dipole moment $ p $, is given by:
$ V = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{p \cos \theta}{r^2} $
Step 2: Identify the given quantities
Dipole moment, $ p = 16 \times 10^{-9} \, \text{C} \cdot \text{m} $
Distance from dipole center, $ r = 0.6 \, \text{m} $
Angle with dipole axis, $ \theta = 60^\circ $
Constant, $ \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{N}\,\text{m}^2/\text{C}^2 $
Step 3: Substitute the values into the formula
Use the fact that $ \cos 60^\circ = \frac{1}{2} $. Hence,
$ V = \left( 9 \times 10^9 \right) \times \frac{ \left(16 \times 10^{-9}\right) \times \cos 60^\circ }{\left(0.6\right)^2} .
$
Step 4: Simplify step by step
Calculate $ p \cos 60^\circ $:
$ 16 \times 10^{-9} \times \frac{1}{2} = 8 \times 10^{-9} \, \text{C} \cdot \text{m} $
Square the distance $ r = 0.6 \, \text{m} $:
$ r^2 = 0.6^2 = 0.36 \, \text{m}^2 $
Multiply by the constant $ \frac{1}{4\pi \varepsilon_0} = 9 \times 10^9 $:
Numerator = $ 9 \times 10^9 \times 8 \times 10^{-9}
= 72 \, (\text{since } 10^9 \times 10^{-9} = 1)
$.
Divide by $ 0.36 $:
$ \frac{72}{0.36} = 200 $
Step 5: State the final answer
Thus, the electric potential at the given point is
$ V = 200 \, \text{V} $.
