Question

A square loop of side $\ell$ having uniform linear charge density $\lambda$ is placed in $xy$ plane as shown in the figure. There is a non-uniform electric field $\overrightarrow{E} = \frac{a}{\ell}(x+\ell)\hat{i}$ where a and $\ell$ are constants. The resultant electric force on the loop is having value $na\lambda\ell$. Find value of $n$.

Answer 1

5

Answer 2

The force on a charged element $dq$ in an electric field $\vec{E}$ is $d\vec{F} = dq \vec{E}$. For a linear charge density $\lambda$, $dq = \lambda dl$. So, $d\vec{F} = \lambda dl \vec{E}$.

The electric field is $\vec{E} = \frac{a}{\ell}(x+\ell)\hat{i}$, always in the $\hat{i}$ direction.

1. Side AB ($x=\ell$, $y$ from $0$ to $\ell$): $\vec{E}=2a\hat{i}$. Force $\vec{F}_{AB} = \int_0^\ell \lambda dy (2a\hat{i}) = 2a\lambda\ell\hat{i}$.

2. Side BC ($y=\ell$, $x$ from $\ell$ to $2\ell$): $\vec{E}=\frac{a}{\ell}(x+\ell)\hat{i}$. Force $\vec{F}_{BC} = \int_{\ell}^{2\ell} \lambda dx \frac{a}{\ell}(x+\ell)\hat{i} = \frac{5}{2}a\lambda\ell\hat{i}$.

3. Side CD ($x=2\ell$, $y$ from $0$ to $\ell$): $\vec{E}=3a\hat{i}$. Force $\vec{F}_{CD} = \int_0^\ell \lambda dy (3a\hat{i}) = 3a\lambda\ell\hat{i}$.

4. Side DA ($y=0$, $x$ from $2\ell$ to $\ell$): $\vec{E}=\frac{a}{\ell}(x+\ell)\hat{i}$. Force $\vec{F}_{DA} = \int_{2\ell}^{\ell} \lambda dx \frac{a}{\ell}(x+\ell)\hat{i} = -\frac{5}{2}a\lambda\ell\hat{i}$.

Summing these forces: $\vec{F}_{total} = (2a\lambda\ell + \frac{5}{2}a\lambda\ell + 3a\lambda\ell - \frac{5}{2}a\lambda\ell)\hat{i} = (2+3)a\lambda\ell\hat{i} = 5a\lambda\ell\hat{i}$.

Comparing with $na\lambda\ell$, we find $n=5$.