Question

Two point charges q, = 1 uC and q, = 2 uC are placed at points A and B 6 cm apart as shownin the figure. A third charge Q = 5 uC is moved from C to D along the arc of a circle of radius8 cm as shown. The change in the potential energy of system is​

Answer 1

Given:

Two point charges q, = 1 uC and q, = 2 uC are placed at points A and B 6 cm apart as shownin the figure.

A third charge Q = 5 uC is moved from C to D along the arc of a circle of radius8 cm as shown.

To find:

The change in the potential energy of system is​

Solution:

From given, we have,

The charge q₁ = 1 × 10^{-6} C

The charge q₂ = 2 × 10^{-6} C

The charge q₃ = 5 × 10^{-6} C

The distance between charges q₁ and q₂ = 6 cm = 0.06 m

The distance between charges q₁ and q₃ = 8 cm = 0.08 m

The distance between charges q₂ and q₃ = 10 cm = 0.1 m

The distance BD = 0.02 m

we use the formula,

ΔU = Uf - Ui

so, we have,

ΔU = 1/4π∈₀ {[(q₁q₃/0.08) + (q₂q₃/0.02)] - [(q₁q₃/0.08) + (q₂q₃/0.1]}

susbstituting the values of charges, we get,

ΔU = 1/4π∈₀ {[(5×10^{-12}/0.08) + (10×10^{-12}/0.02)] - [(5×10^{-12}/0.08) + (10×10^{-12}/0.1]}

ΔU = 9×10^9  {[(10×10^{-12}/0.02)] - [(10×10^{-12}/0.1]}

ΔU = 9 × 10^9 × 10^{-12} [ 10/0.02 - 10/0.1]

ΔU = 9 × 10^{-3} (600)

ΔU = 5.4 J

Therefore, the change in the potential energy of system is​ 5.4 J