Question 28

Two charges $$7\mu c$$ and $$-4\mu c$$ are placed at (−7 cm, 0, 0) and (7 cm, 0, 0) respectively. Given, $$\epsilon_{\circ}=8.85\times 10^{-12}C^{2}m^{-2}$$, the electrostatic potential energy of the charge configuration is :

U = (1 / 4πϵ₀) · (q₁q₂ / r)

given:

q₁ = 7 μC = 7 × 10⁻⁶ C
q₂ = −4 μC = −4 × 10⁻⁶ C

distance between charges:

positions are at −7 cm and +7 cm
so separation r = 14 cm = 0.14 m

step 1: substitute

U = 9 × 10⁹ × (7×10⁻⁶)(−4×10⁻⁶) / 0.14

step 2: simplify

(7×−4) = −28

U = 9 × 10⁹ × (−28 × 10⁻¹²) / 0.14

= 9 × (−28) × 10⁻³ / 0.14

= (−252 × 10⁻³) / 0.14

step 3: final calculation

−0.252 / 0.14 = −1.8

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