Question

Four point charges are placed in a straight line with magnitude and separation as shown in the diagram. What should be the value of q₀ such that + 10µC charge is in equilibrium?

Options

• - 80 µC

• + 40 µC

• + 80 µC

• - 20 µC

Answer 1

+ 80 µC

Explanation:

Force on B due to A (+x) f_AB = `("K""q"_"A""q"_"B")/"r"_"AB"^2`

= `(9xx10^9xx40xx10^-6xx10xx10^-6)/((40xx10^-2)^2) = 90/4`

= 22.5 N

Force B due to C (+x)f_CB = `("K""q"_"C""q"_"B")/"r"_"BC"^2`

= `(9xx10^9xx10xx10^-6xx10xx10^-6)/((20xx10^-2)^2) = 90/4`

= 22.5 N

Force on B due to D (–x) f_BD = `("K""q"_"B""q"_"D")/"r"_"BD"^2`

=  `(9xx10^9xx10xx10^-6xx"q"_0)/((40xx10^-2)^2) = 9/16xx10^6 "q"_0`

For the equilibrium,

f_BD = f_AB + f_CB

⇒ `9/16xx10^6"q"_0` = 22.5 + 22.5

⇒ `9/16xx10^6"q"_0` = 45

⇒ q₀ = `(45xx16)/(9xx10^6)` 

⇒ q₀ = + 80 × 10^-6

⇒ q₀ = + 80 µC