1. Given arrangement

There are $10$ charges placed uniformly on a circle of radius $R$.

• Charges at positions $1,3,5,7,9$ are $+q$
• Charges at positions $2,4,6,8,10$ are $-q$

So the charges alternate in sign around the circle.

2. Potential at the centre

Potential is a scalar quantity, so we add algebraically.

Each charge is at distance $R$ from the centre, so potential due to one charge is

$V_i = \frac{1}{4\pi\varepsilon_0}\frac{q_i}{R}$

Hence total potential at the centre is

$V = \frac{1}{4\pi\varepsilon_0 R}\sum q_i$

Now,

$\sum q_i = 5(+q) + 5(-q) = 0$

Therefore,

$V=0$

3. Electric field at the centre

Electric field is a vector quantity, so direction matters.

Consider any diametrically opposite pair of charges. Since there are $10$ equally spaced charges, each charge has a diametrically opposite partner.

Because the charges alternate in sign, the charge opposite to any given charge is also of opposite sign? Let us check carefully:

• Angular separation between successive charges is $\frac{360^\circ}{10}=36^\circ$
• Diametrically opposite point is $180^\circ$ away, i.e. after $5$ steps

If we move $5$ steps from a $+q$ position, we reach an even position, which has charge $-q$.

So each diametrically opposite pair consists of equal and opposite charges.

Now examine electric field at the centre due to such a pair:

• Field due to $+q$ at the centre is directed away from the positive charge, i.e. along the radius toward the centre from that charge.
• Field due to the opposite $-q$ is directed toward the negative charge, which at the centre is along the same line and in the same direction.

Thus for each diametrically opposite pair, the two field vectors add.

Magnitude due to each charge is

$E_0 = \frac{1}{4\pi\varepsilon_0}\frac{q}{R^2}$

So each opposite pair gives magnitude $2E_0$ along its diameter.

There are $5$ such diameters. Their directions are equally spaced by $36^\circ$ around the circle.

Thus the total field is the vector sum of $5$ equal vectors symmetrically distributed around a full circle. Such a sum is zero.

Equivalently, the vectors are along angles

$0^\circ, 36^\circ, 72^\circ, 108^\circ, 144^\circ$

with equal magnitude, and their vector sum is zero by symmetry.

Hence,

$E=0$

4. Final result

$V=0, \qquad E=0$

So the correct option is A.

5. Comparison with stored answer

Stored correct answer: A

My derived answer: A

They agree.