Question

Five identical charges q are placed on the equidistant on a semicircle as shown
Another point charge q is kept at the center of the circle of radius R. Calculate the electrostatic force

Answer 1

your question is incomplete. A complete question is --------> Five identical point charges, each having charge Q, are equally spaced on a semicircle of radius R as shown in the figure below. Find the force (in terms of k, Q, and R) on a charge q located equidistant from the five other charges. (Use any variable or symbol stated above as necessary.)

solution : force exerted by each Q on q , F = KQq/R²

see figure, force exerted by charge Q located at D is cancelled by force exerted by charge Q located at E .

all five charges are equidistant,

so, angle made by two consecutive charge on centre of it = 90/5 = 18°

so, angle made by charges located at A and B = 2 × 18° = 36°

now, resultant of A and B = √{F² + F² + 2F²cos36°}

= F√{1 + 2cos36°}, it is directed along x - axis .

now, net force experienced by q = force exerted by charge located at C + resultant of charge located at A and B.

= F + F√{1 + 2cos36°}

putting F = KQq/R² ,

now, Fnet = kQq/R²[1 + √(1 + 2cos²36°)