Two charges 7μC and -2 μC are placed at (-9, 0, 0) cm and (9, 0, 0) cm respectively in an external field $\vec{E}$ = (A/r²) r̂, where A = 9 × 10⁵ N/C·m². Considering the potential at infinity is 0, the electrostatic energy of the configuration i $s_____J.$
Options

Question

Two charges 7μC and -2 μC are placed at (-9, 0, 0) cm and (9, 0, 0) cm respectively in an external field $\vec{E}$ = (A/r²) r̂, where A = 9 × 10⁵ N/C·m². Considering the potential at infinity is 0, the electrostatic energy of the configuration i $s_____J.$

Options

G1 Physics · Accepted Answer

The external electric field is E⃗ = (A/r²)r̂. Using dV = -E⃗·dr⃗, and V(∞)=0, V(r) = A/r. Both charges are at distance r = 9 cm = 0.09 m from the origin. External-field contribution to the potential energy is U(ext) = q₁V₁ + q₂V₂ = (q₁+q₂)A/r. Here, q₁+q₂ = 7μC-2μC = 5×10⁻⁶ C. Therefore, U(ext) = (5×10⁻⁶)(9×10⁵)/0.09 = 50 J. The separation between the two charges is d = 18 cm = 0.18 m. Their mutual interaction energy is U(int) = kq₁q₂/d = [9×10⁹×(7×10⁻⁶)(-2×10⁻⁶)]/0.18 = -0.7 J. Hence, U(total)…

Two charges 7μC and -2 μC are placed at (-9, 0, 0) cm and (9, 0, 0) cm respectively in an external field E ⃗ \vec{E} E = (A/r²) r̂, where A = 9 × 10⁵ N/C·m². Considering the potential at infinity is 0, the electrostatic energy of the configuration i s_____J.