In the figure, there are three infinitely long thin sheets having surface charge density + 3σ, -2σ$ and $+σ$ respectively. Determine magnitude and direction of the electric field at point A. |
$\frac{σ}{ε_0}$ towards right $\frac{σ}{2ε_0}$ towards left $\frac{2σ}{ε_0}$ towards right $\frac{σ}{ε_0}$ towards left |
$\frac{σ}{ε_0}$ towards left |
The correct answer is Option (4) → $\frac{σ}{ε_0}$ towards left For an infinite sheet with surface charge density $\sigma$, the electric field on either side is $E = \frac{\sigma}{2 \, \epsilon_0}$ Direction: away from the sheet if $\sigma$ is positive, towards the sheet if $\sigma$ is negative. At point A (to the left of all sheets): From $+3\sigma$ sheet: $E_1 = \frac{3\sigma}{2 \, \epsilon_0}$ towards left From $-2\sigma$ sheet: $E_2 = \frac{2\sigma}{2 \, \epsilon_0} = \frac{\sigma}{\epsilon_0}$ towards right From $+\sigma$ sheet: $E_3 = \frac{\sigma}{2 \, \epsilon_0}$ towards left Net field at A: $E_{\text{net}} = -\frac{3\sigma}{2 \, \epsilon_0} + \frac{\sigma}{\epsilon_0} - \frac{\sigma}{2 \, \epsilon_0}$ $E_{\text{net}} = -\frac{\sigma}{\epsilon_0}$ Therefore, the electric field at A is $\frac{\sigma}{\epsilon_0}$ towards left. |