Question

Two point charges −2μC and 5μC are placed at (−30 cm, 0) and (30 cm, 0) respectively in an external electric field `vecE = A/x^2 hati`, where A = 9 × 10⁵ Nm² C⁻¹. Find the electrostatic potential energy of this configuration.

Answer 1

Given: q₁ =  −2μC

= −2 × 10⁻⁶ C

q₂ = 5μC

= 5 × 10⁻⁶ C

r = |x₂ − x₁|

= |30 − (−30)| cm

= 0.60 cm

= 0.6 m

A = 9 × 10⁵ Nm² C⁻¹

`1/(4pi ε_0) = 9 xx 10^9 N^-1 m^2 C^-2`

Formula: `U_"int" = 1/(4pi ε_0) (q_1 q_2)/r`

= `(4pi ε_0 xx q_1 q_2)/r`

= `(9 xx 10^9 xx (-2 xx 10^-6) xx (5 xx 10^-6))/0.6`

= `(-90 xx 10^-3)/0.6`

= −150 × 10⁻³

= −0.15 J

The relationship between electric field and potential is V = −`intvecE  .  dvecr`

Calculating the potential at each position (A = 9 × 10⁵ Nm²C⁻¹)

E = `A/x^2`

V(x) = `-intA/x^2  dx`

= `A/x`

`V(x_1) = (9 xx 10^5)/(-0.30)`

= −3 × 10⁶ V

`V(x_2) = (9 xx 10^5)/(0.30)`

= 3 × 10⁶ V

In an external field, the energy of each charge in the field is U = qV.

U₁ = q₁V₁

= (−2 × 10⁻⁶) × (−3 × 10⁶)

= 6 J

U₂ = q₂V₂

= (5 × 10⁶) × (3 × 10⁶)

= 15 J

Total Potential Energy 

`U_"total" = U_1 + U_2 + U_"int"`

= 6 + 15 + (−0.15)

= 20.85 J