Question

(i) Two charges of a dipole $- 4 μ C$ and $+ 4 μ C$ are placed at the points $A (1, 0, 4) m$ and $B (2, - 1, 5) m$ located in an electric field $\vec{E} = 0.20 \hat{i} V {cm}^{- 1}$ . Calculate the torque acting on the dipole.

(ii) A Gaussian surface is shown in figure. Charges $q_{1}$ and $q_{2}$ are inside and charge $q_{3}$ is outside the surface. Indicate the charges which contributes the electric field appearing in the formula $\oint \vec{E} \cdot \vec{d s} = \frac{q_{in}}{ε_{0}}$ .

(iii) The electric field in a region is given by $\vec{E} = \frac{3}{5} E_{0} \hat{i} + \frac{4}{5} E_{0} \hat{j}$ with $E_{0} = 2 \times 10^{3} {NC}^{- 1} .$ Find the flux of this field through a rectangular surface of area $0.2 m^{2}$ parallel to $A (1, 0, 4) m$ plane.

Accepted Answer

(i) Given $A (1, 0, 4)$ and $B (2, - 1, 5)$

$∴ 2 \vec{a} = \vec{A B} = [(2 - 1) \hat{i} + (- 1 - 0) \hat{j} + (5 - 4) \hat{k}] = [\hat{i} - \hat{j} + \hat{k}]$

$q = \pm 4 \times 10^{- 6} C$

$E = 0.20 \hat{i} V / c m = 20 \hat{i} V / m$

As we know, torque is given by

$\vec{τ} = \vec{p} \times \vec{E} = q (2 \vec{a}) \times \vec{E}$

$\Rightarrow \vec{τ} = 4 \times 10^{- 6} (\hat{i} - \hat{j} + \hat{k}) \times 20 \hat{i}$

$\Rightarrow \vec{τ} = 8 \times 10^{- 5} (\hat{k} + \hat{j})$

Magnitude of torque,

$τ = 8 \times 10^{- 5} \sqrt{1^{2} + 1^{2}} = 1.13 \times 10^{- 4} N - m$

(ii) According to Gauss' law, the term $q_{e n c l o s e d}$ on the right side of the equation $\oint E . d S = \frac{q_{e n c l o s e d}}{ε_{\circ}}$ includes the sum of all charges enclosed by the surface called (Gaussian surface).

In left side equation, the electric field is due to all the charges present both inside and outside the Gaussian surface.

Hence, in given question, $E$ on L.H.S of the above equation will have a contribution from all charges while $q$ on the R.H.S will have a contribution from $q_{1}$ and $q_{2}$ only.

(iii) The flux is given by, $ϕ = E . \hat{n} d S$

As the area parallel to $y - z$ plane so normal unit vector to surface is $\hat{n} = \hat{i}$

So, $ϕ = (\frac{3}{5} E_{\circ} \hat{i} + \frac{4}{5} E_{\circ} \hat{j}) . \hat{i} (0.2) = \frac{3}{5} E_{\circ} \times 0.2 = \frac{3}{5} \times (2 \times 10^{3}) (0.2) = 240 N . m^{2} / C$

Important Questions on Electric Charges and Fields

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