how to find electric field on a semi charged infinite rod
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HEYA!!
HERE IS YOUR ANSWER,
Well to find the Electric Field on a charged semi infinite long rod :
⇔ E field due to straight wire,
→ Eₓ = λ × (sinθ₂+sinθ₁) / (4πε₀a)
→ Ey = λ × (cosθ₁−cosθ₂) / (4πε₀a)
NOW SINCE,
→ θ₁ = -π/4
→ θ₂ = π/2
THEREFORE,
→ E = [λ × (1−sin(π/4)) ÷ (4πε₀a)] i + [λ × (cos(π/4)) ÷ (4πε₀a)] j (ANS)
HOPE IT HELPS YOU,
THANK YOU. ^_^
HERE IS YOUR ANSWER,
Well to find the Electric Field on a charged semi infinite long rod :
⇔ E field due to straight wire,
→ Eₓ = λ × (sinθ₂+sinθ₁) / (4πε₀a)
→ Ey = λ × (cosθ₁−cosθ₂) / (4πε₀a)
NOW SINCE,
→ θ₁ = -π/4
→ θ₂ = π/2
THEREFORE,
→ E = [λ × (1−sin(π/4)) ÷ (4πε₀a)] i + [λ × (cos(π/4)) ÷ (4πε₀a)] j (ANS)
HOPE IT HELPS YOU,
THANK YOU. ^_^
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