JEE WhatsApp Group
Sign in
Please select an account to continue using cracku.in
↓ →
JEE WhatsApp Group
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
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
Click on the Email ☝️ to Watch the Video Solution
Create a FREE account and get:
Educational materials for JEE preparation
Ask our AI anything
AI can make mistakes. Please verify important information.
AI can make mistakes. Please verify important information.
