In this diagram, three infinite plane sheets of charges are shown with surface charge densities $+σ,+σ, -σ$. The value of electric field at points A and B are |
0 at A and 0 at B $\frac{σ}{2ε_0}$ at A and $\frac{3σ}{2ε_0}$ at B $\frac{3σ}{2ε_0}$ at A and $\frac{σ}{2ε_0}$ at B $\frac{3σ}{2ε_0}$ at A and 0 at B |
$\frac{3σ}{2ε_0}$ at A and $\frac{σ}{2ε_0}$ at B |
The correct answer is Option (3) → $\frac{3σ}{2ε_0}$ at A and $\frac{σ}{2ε_0}$ at B Field of an infinite sheet: magnitude $E=\frac{\sigma}{2\varepsilon_0}$, directed away from $+\sigma$ and toward $-\sigma$. At A (between middle $+\sigma$ and bottom $-\sigma$): Top $+\sigma$: down $\frac{\sigma}{2\varepsilon_0}$; Middle $+\sigma$: down $\frac{\sigma}{2\varepsilon_0}$; Bottom $-\sigma$: down $\frac{\sigma}{2\varepsilon_0}$. Sum $=\frac{3\sigma}{2\varepsilon_0}$ (down). At B (between top $+\sigma$ and middle $+\sigma$): Top $+\sigma$: down $\frac{\sigma}{2\varepsilon_0}$; Middle $+\sigma$: up $\frac{\sigma}{2\varepsilon_0}$ (cancel); Bottom $-\sigma$: down $\frac{\sigma}{2\varepsilon_0}$. Result $=\frac{\sigma}{2\varepsilon_0}$ (down). Answer: $E_A=\frac{3\sigma}{2\varepsilon_0}$ and $E_B=\frac{\sigma}{2\varepsilon_0}$. |