Question

Two point charges 5 μC and −1 μC are placed at points (−3 cm, 0, 0) and (3 cm, 0, 0) respectively. An external electric field `vec E = A/r^2 hat r` where A = 3 × 10⁵ Vm is switched on in the region. Calculate the change in electrostatic energy of the system due to the electric field.

Answer 1

The external electric field is given by:

`vec E = A/r^2 hat r`

The electric potential V(r) is related to the electric field by the expression:

`vec e = -bar V V`

For a radial field:

V(r) = `-int vec E * vec(dr)`

= `-int A/r^2 dr`

= `A/r`

Given A = 3 × 10⁵ Vm, the potential at a distance r from the origin is:

V(r) = `(3 xx 10^5)/r`

The charges are placed at:

q₁ = 5 × 10⁻⁶ C at `vec (r_1)` = (−3 cm, 0, 0).The distance from the origin is:

r₁ = 3 cm = 0.03 m

q₂ = −1 × 10⁻⁶ C at `vec (r_2)` = (3 cm, 0, 0).The distance from the origin is:

r₂ = 3 cm = 0.03 m

Since both charges are at the same radial distance r = 0.03 m from the origin, the potential V at both points is:

V = `(3 xx 10^5)/0.03`

= 10⁷ V

The change in the electrostatic energy of the system when the field is switched on is the energy provided by the external field, calculated as:

ΔU = q₁V(r₁) + q₂V(r₂)

= (5 × 10⁻⁶ C × 10⁷ V) + (−1 × 10⁻⁶ C × 10⁷ V)

= 50 J − 10 J

= 40 J