91Ó°ÊÓ

1. Radius of the circle, r = 8.20cm
2. Charge distributed along one-quarter of the circumference of the circle, Q₁ = +4.20pC
3. Charge distributed along 3/4th of the circumference of the circle, Q₂ = 25.20pC
4. Distance of the point from the center, d = 6.71cm
5. The electric potential at infinity is. V = 0

Using the concept of the electric potential at a point on a thin rod, we can get the individual potential due to each charge. Now, using the sum of these values, we can get the desired values of the potentials at the center and the point considering the distance of the point.

Formulae:

The electric potential due to point charge at a distance r from a point charge is given by, $V=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q}{r}$ (i)

Here, q is magnitude of charge,${\mathrm{Î\mu }}_0$is the permittivity of free space.

The electric potential due to collection of point charges is given by, $V=\underset{i=1}{\overset{n}{∑}}\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q_i}{r_i}$ (ii)

The electric potential V_Q1at the center C of the circle due to charge Q₁ is given using equation (i) as follows:

$V_{Q_1}=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{Q_1}{R}..............\left(a\right)$

Here, R is the radius of circle

The electric potentialV_Q1 at the center C of the circle due to charge Q₁ is given using equation (ii) as follows:

$V_{Q_1}=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{Q_1}{R}...............\left(b\right)$

Now the electrical potential V_center at the center C of the circle due to the charges Q₁ and Q₂ is the sum of the electrical potential due to charge Q₁ and the electrical potential due to chargeQ₂ . That is given using equations (a) and (b) in equation (ii) as follows:

$\begin{array}{rcl}{V}_{center}& =& \frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{Q_1}{R}+\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{Q_2}{R}\\ & =& \frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left(\frac{Q_1}{R}+\frac{Q_2}{R}\right)\\ & & \left(substitutingthegivenvalues\right)\\ & =& 9.0×{10}^{9}N{m}^2/C^2\left(\frac{4.20pC\left({\displaystyle \frac{{10}^{-12}C}{1pC}}\right)}{8.20cm\left(\frac{{10}^{-2}m}{1cm}\right)}+\frac{-6\left(4.20pC\left(\frac{{10}^{-12}C}{1pC}\right)\right)}{8.20cm\left({\displaystyle \frac{{10}^{-2}m}{1cm}}\right)}\right)\\ & =& 9.0×{10}^{9}N{m}^2/C^2\left(\frac{4.20×{10}^{-12}C}{0.082m}+\frac{-25.2×{10}^{-12}C}{0.082m}\right)\\ & =& -2.30V\end{array}$

Hence, the electrical potential V_center at the center C of the circle due to the charges Q₁ and Q₂ is -2.30 V.

From the above figure the r between each charged particle and the point P is given as:

$r=\sqrt{R^2+D^2}$

Now the electric potential at point p due to charge Q₁ is given using equation (i) as follows:

role="math" localid="1662703245948" $V_{Q_1}=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{Q_1}{\sqrt{R^2+D^2}}..................\left(c\right)$

And the electric potential at point P due to charge Q₂ is given using equation (i) as follows:

$V_{Q_2}=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{Q_2}{\sqrt{R^2+D^2}}..................\left(d\right)$

The total net electrical potential V_P at point p due to charge Q₁ and Q₂ is the sum of the electrical potential due to charge Q₁ and the electrical potential due to charge Q₂. That is given using equations (c) and (d) in equation (ii) as follows:

$\begin{array}{rcl}{V}_P& =& \frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{Q_1}{\sqrt{R^2+D^2}}+\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{Q_2}{\sqrt{R^2+D^2}}\\ & =& \frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left(\frac{Q_1+Q_2}{\sqrt{R^2+D^2}}\right)\\ & =& 9.0×{10}^{9}N{m}^2/C^2\left(\frac{4.20×{10}^{-12}C-6\left(4.20×{10}^{-12}C\right)}{\sqrt{{\left(0.082m\right)}^2+{\left(0.067m\right)}^2}}\right)\\ & =& -1.78V\end{array}$

Hence, the total electrical potential at the point P due to charge Q₁ and Q₂ , located at a distance of 6.71 cm is -1.78V.