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Question: Two thin wire rings each having radius \(R\) are placed at a distance \(d\) apart with their axes ...

Two thin wire rings each having radius RR are placed at a distance dd apart with
their axes coinciding. The charges on the two rings are +Q + Q and −Q - Q. The potential
difference between the centres of the two rings is:
A. Zero
B. Q4πε0[1R−1R2+d2]\dfrac{Q}{{4\pi {\varepsilon _0}}}\left[ {\dfrac{1}{R} - \dfrac{1}{{\sqrt {{R^2} + {d^2}} }}} \right]
C. Q4πε0d2\dfrac{Q}{{4\pi {\varepsilon _0}{d^2}}}
D. Q2πε0[1R−1R2+d2]\dfrac{Q}{{2\pi {\varepsilon _0}}}\left[ {\dfrac{1}{R} - \dfrac{1}{{\sqrt {{R^2} + {d^2}} }}} \right]

Explanation

Solution

Use the expression of the electric potential at a point: V=14πε0⋅qrV = \dfrac{1}{{4\pi {\varepsilon _0}}} \cdot \dfrac{q}{r} and the electric potential at the centre of the rings is due to its own
ring and the other ring. Obtain the potential at centre of both rings.
Complete step by step solution:
From the question, we know that the radius of the thin rings is RR, the separation between the
rings is dd, the charge on the thin rings are +Q + Q and −Q - Q.

By using the Pythagoras theorem, we have,
r2=R2+d2 r=R2+d2 {r^2} = {R^2} + {d^2}\\\ r = \sqrt {{R^2} + {d^2}}
Let us find the potential at the centre of the ring of charge +Q + Q,
We know the electric potential develops at the centre of the rings with charge +Q + Q is due to
the potential of the ring itself and the electric potential due to the other ring with charge −Q - Q, So,

V1=14πε0⋅QR+14πε0⋅(−Q)r{V_1} = \dfrac{1}{{4\pi {\varepsilon _0}}} \cdot \dfrac{Q}{R} + \dfrac{1}{{4\pi {\varepsilon _0}}} \cdot \dfrac{{\left( { - Q} \right)}}{r}
We can write the above equation as,
V1=14πε0[QR−QR2+d2]{V_1} = \dfrac{1}{{4\pi {\varepsilon _0}}}\left[ {\dfrac{Q}{R} - \dfrac{Q}{{\sqrt {{R^2} + {d^2}} }}} \right] … (I)
Similarly, the electric potential develops at the centre of the rings with charge −Q - Q is due to
the potential of the ring itself and the electric potential due to the other ring with charge +Q + Q, So,
V2=14πε0⋅Qr+14πε0⋅(−Q)R{V_2} = \dfrac{1}{{4\pi {\varepsilon _0}}} \cdot \dfrac{Q}{r} + \dfrac{1}{{4\pi {\varepsilon _0}}} \cdot \dfrac{{\left( { - Q} \right)}}{R}
We can write the above equation as,

\dfrac{Q}{R}} \right]$$ … (II) Now the potential difference at the centres of the two rings is calculated as, $\Delta V = {V_1} - {V_2}$ Substitute the expression in the above equation, we have, $$\Delta V = \dfrac{1}{{4\pi {\varepsilon _0}}}\left[ {\dfrac{Q}{R} - \dfrac{Q}{{\sqrt {{R^2} + {d^2}} }}} \right] - \dfrac{1}{{4\pi {\varepsilon _0}}}\left[ {\dfrac{Q}{{\sqrt {{R^2} + {d^2}} }} - \dfrac{Q}{R}} \right]$$ Now we rewrite and simplify the above equation,

\Delta V = \dfrac{1}{{4\pi {\varepsilon _0}}}\left[ {\dfrac{Q}{R} - \dfrac{Q}{{\sqrt {{R^2} + {d^2}}
}} - \dfrac{Q}{{\sqrt {{R^2} + {d^2}} }} + \dfrac{Q}{R}} \right]\\
\Delta V = \dfrac{1}{{4\pi {\varepsilon _0}}}\left[ {\dfrac{{2Q}}{R} - \dfrac{{2Q}}{{\sqrt {{R^2} +
{d^2}} }}} \right]

After further simplification, we get, $\Delta V = \dfrac{Q}{{2\pi {\varepsilon _0}}}\left[ {\dfrac{1}{R} - \dfrac{1}{{\sqrt {{R^2} + {d^2}} }}} \right]$ **Thus, the option (D) is correct.** **Note:** The electric potential at a point in an electric field is the work needed to move a unit charge (positive) from infinity to that point against the electrostatic forces and it is spherically symmetric as it depends only on the distance of the point from the charge.