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Question

(B) For two infinitely large parallel conducting plates,the charge redistributes such that the outer surfaces have equal charge density $\sigma_{out}

Vedclass · Accepted Answer

$\frac{3 \sigma q}{2 \epsilon_0}$

Vedclass · Answer

(B) For two infinitely large parallel conducting plates,the charge redistributes such that the outer surfaces have equal charge density $\sigma_{out} = \frac{\sigma_1 + \sigma_2}{2} = \frac{\sigma + (-2\sigma)}{2} = -\frac{\sigma}{2}$.The inner surfaces will have charge densities $\sigma_{in1} = \sigma - (-\frac{\sigma}{2}) = \frac{3\sigma}{2}$ and $\sigma_{in2} = -2\sigma - (-\frac{\sigma}{2}) = -\frac{3\sigma}{2}$.The electric field between the plates is due to the inner surface charges: $E = \frac{\sigma_{in1}}{2\epsilon_0} + \frac{|\sigma_{in2}|}{2\epsilon_0} = \frac{3\sigma/2}{2\epsilon_0} + \frac{3\sigma/2}{2\epsilon_0} = \frac{3\sigma}{2\epsilon_0}$.The force experienced by the charge $+q$ is $F = qE = q \left( \frac{3\sigma}{2\epsilon_0} \right) = \frac{3\sigma q}{2\epsilon_0}$.

Consider two infinitely large plane parallel conducting plates as shown below. The plates are uniformly charged with a surface charge density $+\sigma$ and $-2 \sigma$. The force experienced by a point charge $+q$ placed at the midpoint between the two plates will be:

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