Two thin wire rings each having a radius R are placed at a distan… - Vidyadip

MCQ

Two thin wire rings each having a radius $R$ are placed at a distance $d$ apart with their axes coinciding. The charges on the two rings are $ + q$ and $ - q$. The potential difference between the centres of the two rings is

A
Zero
B
$\frac{Q}{{4\pi {\varepsilon _0}}}\,\left[ {\frac{1}{R} - \frac{1}{{\sqrt {{R^2} + {d^2}} }}} \right]$
C
$QR/4\pi {\varepsilon _0}{d^2}$
✓
$\frac{Q}{{2\pi {\varepsilon _0}}}\left[ {\frac{1}{R} - \frac{1}{{\sqrt {{R^2} + {d^2}} }}} \right]$

✓

Answer

Correct option: D.

$\frac{Q}{{2\pi {\varepsilon _0}}}\left[ {\frac{1}{R} - \frac{1}{{\sqrt {{R^2} + {d^2}} }}} \right]$

d
(d) Potential at the centre of rings are
${V_{{O_1}}} = \frac{{k.q}}{R} + \frac{{k( - q)}}{{\sqrt {{R^2} + {d^2}} }}$, ${V_{{O_2}}} = \frac{{k( - q)}}{R} + \frac{{kq}}{{\sqrt {{R^2} + {d^2}} }}$
$==>$ ${V_{{O_1}}} - {V_{{O_2}}} = 2kq\,\left[ {\frac{1}{R} - \frac{1}{{\sqrt {{R^2} + {d^2}} }}} \right]$$ = \frac{q}{{2\pi {\varepsilon _0}}}\left[ {\frac{1}{R} - \frac{1}{{\sqrt {{R^2} + {d^2}} }}} \right]$

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