Question

Five identical charges, Q = 2μC are placed equidistant on a semicircle as shown in the figure. Another point charge q = 1μC is kept at the center of the circle of radius 2 cm. Calculate the electrostatic force experienced by the charge q.

Answer 1

Force on q due to any one charge Q at distance R is

F₀ = `1/(4 pi epsilon_0) (qQ)/R^2`

Charges at the top and bottom (symmetric about the horizontal diameter) produce equal and opposite forces on q; hence, they cancel.

The two charges at 45° above and below the horizontal are symmetric: their vertical components cancel and their horizontal components add.

`"F"_"(pair)"​ = 2F_0 ​cos 45° = sqrt2 F_0`

The middle charge on the left gives a force of F₀​ along the horizontal.

Therefore, resultant force on q is

F = `F_0 + sqrt2 F_0 = F_0 (1 + sqrt2)`

F = `1/(4 pi epsilon_0) (q Q)/R^2 (1 + sqrt2)`