Six point charges are kept 60° apart from each other on the circumference of a circle of radius R as shown in figure. The net electric field at the center of the circle is ____ .
( ε₀ is permittivity of free space)

Solution

Physics sir · 4 steps

Pair up opposite charges: equal ones cancel, and only the +Q and −Q pair survives.

1. Place the charges. Measure angles from the +x axis. The +Q charges sit at 30°, 90°, 210°, 270° and 330°, and the -Q charge sits at 150°. Each one is a distance R from the centre, so each field has magnitude E = Q/(4πε₀ R²).
2. Cancel equal pairs. The +Q at 90° and the +Q at 270° are diametrically opposite, so their fields at the centre are equal and opposite. The same is true for the +Q charges at 30° and 210°. Both pairs give zero.
3. The surviving pair. The field of the +Q at 330° points away from it, towards 150°. The field of the -Q at 150° points towards it, also towards 150°. These two add, giving 2E along the unit vector (-√3/2î + 1/2ĵ).

   E⃗ = 2·Q/(4πε₀ R²)(-√3/2î + 1/2ĵ) = -Q/(4πε₀ R²)(√3 î - ĵ)

4. So. The net field at the centre is -Q/(4πε₀ R²)(√3 î - ĵ), which is option (c).

The idea to keep. For charges arranged symmetrically on a circle, first cancel equal charges sitting opposite each other. An opposite pair of +Q and −Q does not cancel. Their fields add, giving double the field.

Answer from the NTA final answer key. Solution by PrepWiser.