Step 1: Write the formula for electric potential

The electric potential $\mathrm{V}$ at a distance $\mathrm{r}$ from a charge $\mathrm{q}$ is:

$$ \mathrm{V} = \frac{1}{4 \pi \varepsilon_0} \frac{q}{r} $$

Step 2: Find the potential at point B ($\mathrm{V}_B$)

At B, the distance from $+q$ is $(x+y)$ and from $-q$ is $x$. So, $$ \mathrm{V}_B = \frac{1}{4\pi\varepsilon_0} \frac{q_1}{x+y} - \frac{1}{4\pi\varepsilon_0} \frac{q_2}{x} $$

Step 3: Find the potential at point A ($\mathrm{V}_A$)

At A, the distance from $+q$ is $x$ and from $-q$ is $(x+y)$. So, $$ \mathrm{V}_A = \frac{1}{4\pi\varepsilon_0} \frac{q_1}{x} - \frac{1}{4\pi\varepsilon_0} \frac{q_2}{x+y} $$

Step 4: $q_1$ and $q_2$ values

Here, $q_1 = +q$ and $q_2 = -q$.

Step 5: Subtract the potentials

Find $\mathrm{V}_B - \mathrm{V}_A$: $$ \begin{aligned} \mathrm{V}_B - \mathrm{V}_A &= \left[ \frac{1}{4\pi\varepsilon_0} \frac{q_1}{x+y} - \frac{1}{4\pi\varepsilon_0} \frac{q_2}{x} \right] - \left[ \frac{1}{4\pi\varepsilon_0} \frac{q_1}{x} - \frac{1}{4\pi\varepsilon_0} \frac{q_2}{x+y} \right] \\ &= \frac{1}{4\pi\varepsilon_0} \left( \frac{q_1}{x+y} - \frac{q_2}{x} - \frac{q_1}{x} + \frac{q_2}{x+y}\right ) \\ &= \frac{1}{4\pi\varepsilon_0} \left( \frac{q_1}{x+y} + \frac{q_2}{x+y} - \frac{q_1}{x} - \frac{q_2}{x} \right) \end{aligned} $$

Step 6: Substitute the values of $q_1$ and $q_2$

Put $q_1 = +q$, $q_2 = -q$:

$$ \frac{1}{4\pi\varepsilon_0} \left( \frac{q-q}{x+y} - \frac{q-(-q)}{x} \right ) = \frac{1}{4\pi\varepsilon_0} \left(0 - \frac{2q}{x}\right) $$

There is a mistake here; let's rewrite the combined terms in the full equation with the charges substituted:

$$ \mathrm{V}_B = \frac{1}{4\pi\varepsilon_0} \left(\frac{q}{x+y} - \frac{q}{x}\right) $$

$$ \mathrm{V}_A = \frac{1}{4\pi\varepsilon_0} \left(\frac{q}{x} - \frac{q}{x+y}\right) $$

So,

$$ \mathrm{V}_B - \mathrm{V}_A = \frac{1}{4\pi\varepsilon_0} \left( \frac{q}{x+y} - \frac{q}{x} - \frac{q}{x} + \frac{q}{x+y} \right ) $$

$$ = \frac{1}{4\pi\varepsilon_0} \left( 2\left(\frac{q}{x+y}\right) - 2\left(\frac{q}{x}\right) \right ) $$

$$ = \frac{2q}{4\pi\varepsilon_0} \left ( \frac{1}{x+y} - \frac{1}{x} \right ) $$

$$ = \frac{2q}{4\pi\varepsilon_0} \left( \frac{x - (x+y)}{x(x+y)} \right ) $$

$$ = \frac{2q}{4\pi\varepsilon_0} \left( \frac{-y}{x(x+y)} \right ) $$

$$ = -\frac{2q y}{4\pi\varepsilon_0 x (x+y)} $$

Step 7: Proportionality

This shows: $$ (\mathrm{V}_A - \mathrm{V}_B) \propto \frac{2q y}{x (x+y)} $$ So, the difference in potential depends on $q$, $y$, $x$, and $(x+y)$ in this way.

Electrostatics

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