An electric dipole is placed as shown in the figure. The electric potential (in 10² V) at the point P due to t

Question

An electric dipole is placed as shown in the figure. The electric potential (in 10² V) at the point P due to the dipole is: (ε₀ = permittivity of free space and 1/4πε₀ = k)

Accepted Answer

1. **Principle of Superposition:** The electric potential ($V$) at any point due to a system of charges is the algebraic sum of the potentials due to the individual charges. For a dipole consisting of charges $+q$ and $-q$, the potential at point P is given by: 
   $V_P = V_{+q} + V_{-q} = k \frac{q}{r_1} + k \frac{-q}{r_2} = kq \left( \frac{1}{r_1} - \frac{1}{r_2} \right)$, where $r_1$ and $r_2$ are the distances of point P from $+q$ and $-q$ respectively.
2. **Analysis of Probable Answer:** The correct option is given as $\frac{3}{8}qk$. This implies that the geometric factor $\left( \frac{1}{r_1} - \frac{1}{r_2} \right)$ must equal $\frac{3}{8}$. 
3. **Hypothetical Geometry:** For example, if the distance from $+q$ to P is 2 units ($r_1 = 2$) and from $-q$ to P is 8 units ($r_2 = 8$), then $\frac{1}{2} - \frac{1}{8} = rac{4}{8} - rac{1}{8} = rac{3}{8}$. Without the figure, the exact distances cannot be verified, but the method remains the scalar addition of potentials.

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