Question 833: Electrostats & Capacitance

Question

Four point charges each +q is placed on the circumference of a circle of diameter 2 d in such a way that they form a square. The potential at the centre is proportional to

Accepted Answer

Four point charges, each with a charge of $ +q $, are positioned on the circumference of a circle with a diameter of $ 2d $, arranged to form a square. To determine the potential at the circle's center, consider the following:
Since the charges are equally distributed on the circle's circumference, the distance from each charge to the center is:
$ r = \frac{	ext{Diameter}}{2} = \frac{2d}{2} = d $
The potential at the center from each charge can be determined using the formula for the potential due to a point charge, $ V = \frac{1}{4\pi \varepsilon_0} \frac{q}{r} $. Therefore, the total potential at the center due to all four charges is:
$ \begin{aligned} V_{	ext{total}} & = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{d} + \frac{q}{d} + \frac{q}{d} + \frac{q}{d} ight) \ & = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{4q}{d} \ & = 4K \left( \frac{q}{d} ight) \end{aligned} $
where $ K = \frac{1}{4 \pi \varepsilon_0} $.
Thus, the potential at the center is proportional to $ \frac{q}{d} $.

Answer

$\frac{\mathrm{q}^2}{\mathrm{~d}^2}$

Answer

$\frac{\mathrm{q}}{\mathrm{d}}$

Answer

$\mathrm{\frac{d}{q}}$

Answer

$\frac{\mathrm{d}^2}{\mathrm{q}^2}$

Electrostats & Capacitance

Four point charges each +q is placed on the circumference of a circle of diameter 2 d in such a way that they form a square. The potential at the centre is proportional to

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