Electric potential due to a space is given by $\phi(x, y, z)=\phi_0 \frac{x_0}{x}$; when $x_0=5 \mathrm{~m}$…
Electric potential due to a space is given by $\phi(x, y, z)=\phi_0 \frac{x_0}{x}$; when $x_0=5 \mathrm{~m}$ and $\phi_0=8 V$. Find the electric field at $(10 \mathrm{~m}, 5 \mathrm{~m}, 5 \mathrm{~m})$
$0.40 \mathrm{Vm}^{-1} \hat{\mathrm{i}}$
$-0.40 \mathrm{Vm}^{-1} \hat{\mathrm{i}}$
$4.0 \mathrm{Vm}^{-1} \hat{\mathrm{i}}$
$-4.0 \mathrm{Vm}^{-1} \hat{\mathrm{i}}$
Solution
We have
$E=-\frac{d v}{d x}=-\frac{d}{d x} \frac{\left[\phi_0 x_0\right]}{x}=-\phi_0 x_0 \times-\frac{1}{x^2}=\frac{40}{x^2}$
So, $\left.\mathrm{E}\right|_{(10,5,5)}=\frac{40}{10^2}=0.4 \mathrm{~V} / \mathrm{m}$