An infinite sheet carrying a uniform surface | vedclass

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Question

(A) The electric field due to an infinite sheet of charge with surface charge density $\sigma$ is given by $\vec{E} = \frac{\sigma}{2\varepsilon_{0}}

Vedclass · Accepted Answer

$\frac{3 \sigma a q}{2 \varepsilon_{0}}$

Vedclass · Answer

(A) The electric field due to an infinite sheet of charge with surface charge density $\sigma$ is given by $\vec{E} = \frac{\sigma}{2\varepsilon_{0}} \hat{k}$ for $z > 0$ and $\vec{E} = -\frac{\sigma}{2\varepsilon_{0}} \hat{k}$ for $z Work done $W = -q \int_{A}^{B} \vec{E} \cdot d\vec{r} = -q \vec{E} \cdot (\vec{r}_{B} - \vec{r}_{A})$.Given $\vec{r}_{A} = a(\hat{i} + 2\hat{j} + 3\hat{k})$ and $\vec{r}_{B} = a(\hat{i} - 2\hat{j} + 6\hat{k})$.Displacement $\vec{d} = \vec{r}_{B} - \vec{r}_{A} = a(\hat{i} - \hat{i}) + a(-2\hat{j} - 2\hat{j}) + a(6\hat{k} - 3\hat{k}) = a(-4\hat{j} + 3\hat{k})$.Work done $W = -q (\frac{\sigma}{2\varepsilon_{0}} \hat{k}) \cdot a(-4\hat{j} + 3\hat{k}) = -\frac{q \sigma a}{2\varepsilon_{0}} (\hat{k} \cdot -4\hat{j} + \hat{k} \cdot 3\hat{k}) = -\frac{q \sigma a}{2\varepsilon_{0}} (0 + 3) = -\frac{3q \sigma a}{2\varepsilon_{0}}$.Note: The magnitude of work done is $\frac{3q \sigma a}{2\varepsilon_{0}}$.

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