Question

If a charge of 1 µC is placed at the origin and another charge of 3 µC is placed at the point (20 m, 0 m, 0 m) in an external uniform electric field of `40 V/m hat i` with the electric potential at the origin to be zero. Find the electrical potential energy of the system.

Answer 1

Given:

q₁ = 1 µC = 1 × 10⁻⁶ C

q₂ = 3 µC = 3 × 10⁻⁶ C

The charge q₁ is placed at the origin (0, 0, 0), while the charge q₂ is located at the point (20 m, 0, 0).

External uniform electric field:

`vecE = 40 V/m hati`

Electric potential at origin = 0  ....(So reference for external field potential is fixed.)

Distance between the charges (r₁₂) = 20 m

1. Potential energy due to interaction between the charges:

`"U"_"charges" = (K q_1 q_2)/r_12`

= `((9 xx 10^9) xx (1 xx 10^-6) xx (3 xx 10^-6))/20`

= `(27 xx 10^-3)/20`

= 1.35 × 10⁻³ J

2. Potential energy due to external electric field:

Potential at a point in a uniform field:

V = `-vecE * vecr`

At origin, V(0) = 0

Thus potential at x = 20 m

V(20) = −40 × 20

= −800 V

Energy of charge q₂​ in this potential:

U_field = q₂ V(20)

= 3 × 10⁻⁶ × (−800)

= −2.4 × 10⁻³ J

3. Total electrical potential energy:

`"U"_"total" = "U"_"charges" + "U"_"field"`

= (1.35 × 10⁻³) + (−2.4 × 10⁻³)

= −1.05 × 10⁻³ J