If the three charges +q, +2q and -q are placed at points A, B, and C respectively, then find the ratio of the net force on the charge at A to the charge at B.

CONCEPT:

Coulomb's law in Electrostatics:

• It state’s that the force of interaction between two stationary point charges is directly proportional to the product of the charges, and inversely proportional to the square of the distance between them and acts along the straight line joining the two charges.

⇒ F ∝ q1 × q2

\(⇒ F \propto \;\frac{1}{{{r^2}}}\)

\(⇒ F = k\frac{{{q_1}\; \times \;{q_2}}}{{{r^2}}}\)

Where k = constant called electrostatic force constant.

• The value of K depends on the nature of the medium between the two charges and the system of units chosen

For vacuum or air,

\(⇒ k=\frac{1}{4\piϵ_{o}}\)    

For medium,

\(⇒ k=\frac{1}{4\piϵ}\)   

\(⇒ \epsilon =\epsilon_{o} \epsilon_{r}\)

Where q1 and q2 = charges, r = distance between the charges, ϵo = permittivity of vacuum, ϵ = permittivity of the medium, and ϵr = dielectric constant of the medium

CALCULATION:

Given q_A = +q, q_B = +2q, q_C = -q, and AB = BC = x

• According to Coulomb's law, the force between the charge at A and the charge at B is given as,

\(⇒ F_{AB} = k\frac{{{q_A}\; \times \;{q_B}}}{{{AB^2}}}\)

\(⇒ F_{AB} = k\frac{{{q}\; \times \;{2q}}}{{{x^2}}}\)

\(⇒ F_{AB} = 2\frac{kq^2}{{{x^2}}}\)     -----(1)

• According to Coulomb's law, the force between the charge at A and the charge at C is given as,

\(⇒ F_{AC} = k\frac{{{q_A}\; \times \;{q_C}}}{{{AC^2}}}\)

\(⇒ F_{AC} = k\frac{{{q}\; \times \;{q}}}{{{(2x)^2}}}\)

\(⇒ F_{AC} = \frac{kq^2}{{{4x^2}}}\)     -----(2)

• According to Coulomb's law, the force between the charge at B and the charge at C is given as,

\(⇒ F_{BC} = k\frac{{{q_C}\; \times \;{q_B}}}{{{CB^2}}}\)

\(⇒ F_{BC} = k\frac{{{q}\; \times \;{2q}}}{{{x^2}}}\)

\(⇒ F_{BC} = 2\frac{kq^2}{{{x^2}}}\)     -----(3)

• The net force on the charge at A is given as,

⇒ F_A = F_AB - F_AC

\(⇒ F_{A} = 2\frac{kq^2}{{{x^2}}}-\frac{kq^2}{{{4x^2}}}\)

\(⇒ F_{A} = \frac{7kq^2}{{{4x^2}}}\)     -----(4)

• The net force on the charge at B is given as,

⇒ FB = FAB + FBC

\(⇒ F_{B} = 2\frac{kq^2}{{{x^2}}}+2\frac{kq^2}{{{x^2}}}\)

\(⇒ F_{B} = \frac{4kq^2}{{{x^2}}}\)     -----(5)

By equation 4 and equation 5,

\(⇒ \frac{F_{A}}{F_B} = \frac{7kq^2}{{{4x^2}}}\times \frac{{{x^2}}}{4kq^2}\)

\(⇒ \frac{F_{A}}{F_B} = \frac{7}{16}\)

• Hence, option 1) is correct.