Govt. Exams
Using Gauss's law with cylindrical surface: E × 2πr = λL/ε₀, hence E = λ/(2πε₀r)
After contact: charge on each = (Q - 3Q)/2 = -Q. Force F = k(-Q)(-Q)/r² = kQ²/r², attractive (opposite signs).
By symmetry, radial components cancel. Axial component: E = kQx/(a² + x²)^(3/2)
Connected spheres have same potential. V = kQ/r, so Q ∝ r. Both statements are equivalent and correct.
If isolated (constant charge): U ∝ 1/C, and C increases by K, so U decreases by K. If connected to battery (constant V): U ∝ C, increases by K. Given 'charged' implies isolated.
For zero potential: 8/(x) = 2/(3-x). Solving: 8(3-x) = 2x → 24 = 10x → x = 2.4 m
Potential energy of dipole: U = -p·E = -pE cosθ, where θ is angle between p and E. Minimum at θ = 0
Capacitance of isolated sphere: C = 4πε₀R, where R is radius. Since k = 1/(4πε₀), C = R/k
By Gauss's law, field outside depends only on total enclosed charge Q. E = kQ/r² for r > R₂
Electrostatic force is conservative, meaning work depends only on initial and final positions, not the path taken
