Question

Ten charges are placed on the circumference of a circle of the radius $R$ with constant angular separation between successive charges. Alternate charges $1$, $3$, $5$, $7$, $9$ have charge (+q) each, while $2$, $4$, $6$, $8$, $10$ have charge (–q) each. The potential $V$ and the electric field $E$ at the center of the circle are respectively: (Take $V=0$ at infinity)

• A. $V=0,E=0$
• B. $V=10q/\left[4{\mathrm{\pi \epsilon }}_0\mathrm{R}\right],E=\raisebox{1ex}{$10q$}\!\left/ \!\raisebox{-1ex}{$\left[4{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^2\right]$}\right.$
• C. $V=0,E=\raisebox{1ex}{$10q$}\!\left/ \!\raisebox{-1ex}{$\left[4{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^2\right]$}\right.$
• D. $V=\raisebox{1ex}{$10q$}\!\left/ \!\raisebox{-1ex}{$\left[4{\mathrm{\pi \epsilon }}_0\mathrm{R}\right],E=0$}\right.$

Answer 1

The correct option is A

$V=0,E=0$

Explanation:

Potential $V$:

Step 1.Given data:

1. As mentioned in the question, charges are placed at constant angular separation. i.e, the angle between them is fixed. then the angle will be $=\raisebox{1ex}{$360$}\!\left/ \!\raisebox{-1ex}{$10$}\right.=36°$
2. Charge on $1,3,5,7,9=+q$
3. Charge on $2,4,6,8,10=-q$

Step 2. Explanation:

1. Potential at any point at a distance $R$ from a charge $q$ can be given as, $V=\frac{kq}{R}$, where $k=\frac{1}{4{\mathrm{\pi \epsilon }}_0}$, ${\epsilon }_0=$permittivity of free space, $q=$charge, $R=$distance.
2. Hence the potential at the centre of the circle due to $+q$ charge is $V=+\frac{kq}{R}$.
3. We have five $+q$ charges, so the combined potential at the centre due to those five positive charges, $V_{+}=+\frac{5kq}{R}$.
4. Similarly, in the case of five negative charges, the combined potential at the centre is, $V_{-}=\frac{-5kq}{R}$
5. Now, Net Potential $V_{net}=V_{+}+V_{-}=5\raisebox{1ex}{$kq$}\!\left/ \!\raisebox{-1ex}{$R$}\right.+5\raisebox{1ex}{$k\left(-q\right)$}\!\left/ \!\raisebox{-1ex}{$R$}\right.=0$

Electric field $E$:

Step 1.Given data:

Charges placed on positions one and six will form a vector as shown in the above figure.

Charges placed on positions two and seven will form a vector as shown in the above figure.

Charges placed on positions three and eight will form a vector as shown in the above figure.

Charges placed on positions four and nine will form a vector as shown in the above figure.

Charges placed on positions five and ten will form a vector as shown in the above figure.

Thus, the combined figure will look like the below shown figure:

1. We can readily see that every vector shown in figure is a combination of the electric field due to one positive charge and one negative charge.
2. Now, all these charges are forming a cyclic vector in the center.

Step 2. Explanation:

1. We know, that if five vectors are acting at a point, the angle between these vectors is equal, and the net resultant is a null vector, which means ${\overrightarrow{R}}_{net}=0$
2. The magnitude of electric field at the center due to one charge = $E=\frac{kq}{R^2}$, where $k=\frac{1}{4{\mathrm{\pi \epsilon }}_0}$, ${\epsilon }_0=$the permittivity of free space, $q=$charge, $R=$distance.
3. Every vector denoted by $E$ at the center represents the influence of two charges. the angle between any two vectors will be $36+36=72°$.
4. The magnitude of every single vector at the center $=\frac{2kq}{R^2}$
5. With the help of all five cyclic vectors in the center, we can form a polygon.
6. And hence the resultant of a cyclic vector is always zero.$E_{net}=0$

Hence option A is the correct option.