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Question: external uniform electric field parallel to x-axis is applied such that net electric field at A is a...

external uniform electric field parallel to x-axis is applied such that net electric field at A is along OA‾\overline{OA}. The magnitude of net electric field at point A is. (K=14πϵ0)\left(K = \frac{1}{4\pi\epsilon_0}\right)

A

kpr3\frac{kp}{r^3}

B

kpcos⁡θr3\frac{kp\cos\theta}{r^3}

C

3kpr3\frac{3kp}{r^3}

D

3kpcos⁡θr3\frac{3kp\cos\theta}{r^3}

Answer

3kpcos⁡θr3\frac{3kp\cos\theta}{r^3}

Explanation

Solution

The electric field at point A due to a dipole with dipole moment p⃗\vec{p} at the origin is given by E⃗dipole=Kr3[3(p⃗⋅r^)r^−p⃗]\vec{E}_{dipole} = \frac{K}{r^3}[3(\vec{p}\cdot\hat{r})\hat{r} - \vec{p}], where r^\hat{r} is the unit vector along OA‾\overline{OA}.

In this case, the dipole is along the x-axis, so p⃗=pi^\vec{p} = p\hat{i}. The position vector of A is r⃗=r(cos⁡θi^+sin⁡θj^)\vec{r} = r(\cos\theta\hat{i} + \sin\theta\hat{j}), and r^=cos⁡θi^+sin⁡θj^\hat{r} = \cos\theta\hat{i} + \sin\theta\hat{j}.

p⃗⋅r^=(pi^)⋅(cos⁡θi^+sin⁡θj^)=pcos⁡θ\vec{p}\cdot\hat{r} = (p\hat{i})\cdot(\cos\theta\hat{i} + \sin\theta\hat{j}) = p\cos\theta.

E⃗dipole=Kr3[3(pcos⁡θ)(cos⁡θi^+sin⁡θj^)−pi^]=Kpr3[3cos⁡2θi^+3sin⁡θcos⁡θj^−i^]\vec{E}_{dipole} = \frac{K}{r^3}[3(p\cos\theta)(\cos\theta\hat{i} + \sin\theta\hat{j}) - p\hat{i}] = \frac{Kp}{r^3}[3\cos^2\theta\hat{i} + 3\sin\theta\cos\theta\hat{j} - \hat{i}].

E⃗dipole=Kpr3[(3cos⁡2θ−1)i^+3sin⁡θcos⁡θj^]\vec{E}_{dipole} = \frac{Kp}{r^3}[(3\cos^2\theta - 1)\hat{i} + 3\sin\theta\cos\theta\hat{j}].

An external uniform electric field parallel to the x-axis is applied, let it be E⃗ext=E0i^\vec{E}_{ext} = E_0\hat{i}.

The net electric field at A is E⃗net=E⃗dipole+E⃗ext=Kpr3[(3cos⁡2θ−1)i^+3sin⁡θcos⁡θj^]+E0i^\vec{E}_{net} = \vec{E}_{dipole} + \vec{E}_{ext} = \frac{Kp}{r^3}[(3\cos^2\theta - 1)\hat{i} + 3\sin\theta\cos\theta\hat{j}] + E_0\hat{i}.

E⃗net=(Kpr3(3cos⁡2θ−1)+E0)i^+(Kpr3(3sin⁡θcos⁡θ))j^\vec{E}_{net} = \left(\frac{Kp}{r^3}(3\cos^2\theta - 1) + E_0\right)\hat{i} + \left(\frac{Kp}{r^3}(3\sin\theta\cos\theta)\right)\hat{j}.

We are given that the net electric field at A is along OA‾\overline{OA}, which means it is along the direction of r^=cos⁡θi^+sin⁡θj^\hat{r} = \cos\theta\hat{i} + \sin\theta\hat{j}.

So, E⃗net=Enetr^=Enet(cos⁡θi^+sin⁡θj^)\vec{E}_{net} = E_{net}\hat{r} = E_{net}(\cos\theta\hat{i} + \sin\theta\hat{j}).

Comparing the components of the two expressions for E⃗net\vec{E}_{net}:

i-component: Kpr3(3cos⁡2θ−1)+E0=Enetcos⁡θ\frac{Kp}{r^3}(3\cos^2\theta - 1) + E_0 = E_{net}\cos\theta

j-component: Kpr3(3sin⁡θcos⁡θ)=Enetsin⁡θ\frac{Kp}{r^3}(3\sin\theta\cos\theta) = E_{net}\sin\theta

From the j-component equation, assuming sin⁡θ≠0\sin\theta \neq 0, we can divide by sin⁡θ\sin\theta:

Enet=Kpr3(3cos⁡θ)=3Kpcos⁡θr3E_{net} = \frac{Kp}{r^3}(3\cos\theta) = \frac{3Kp\cos\theta}{r^3}.

Now substitute this expression for EnetE_{net} into the i-component equation:

Kpr3(3cos⁡2θ−1)+E0=(3Kpcos⁡θr3)cos⁡θ=3Kpcos⁡2θr3\frac{Kp}{r^3}(3\cos^2\theta - 1) + E_0 = \left(\frac{3Kp\cos\theta}{r^3}\right)\cos\theta = \frac{3Kp\cos^2\theta}{r^3}.

E0=3Kpcos⁡2θr3−Kpr3(3cos⁡2θ−1)=Kpr3(3cos⁡2θ−(3cos⁡2θ−1))=Kpr3(3cos⁡2θ−3cos⁡2θ+1)=Kpr3E_0 = \frac{3Kp\cos^2\theta}{r^3} - \frac{Kp}{r^3}(3\cos^2\theta - 1) = \frac{Kp}{r^3}(3\cos^2\theta - (3\cos^2\theta - 1)) = \frac{Kp}{r^3}(3\cos^2\theta - 3\cos^2\theta + 1) = \frac{Kp}{r^3}.

So the magnitude of the external electric field is E0=Kpr3E_0 = \frac{Kp}{r^3}.

The magnitude of the net electric field at point A is Enet=3Kpcos⁡θr3E_{net} = \frac{3Kp\cos\theta}{r^3}.