Electrostatics Quiz

$E = \frac{1}{4\pi\varepsilon_0} \left(\frac{q r}{R^3}\right)$, Here q is the total charge (correct)

$E = \frac{1}{4\pi\varepsilon_0} \left(\frac{q r}{R^2}\right)$, Here q is the total charge

$E = \frac{1}{4\pi\varepsilon_0} \left(\frac{q}{R^2}\right)$, Here q is the total charge

$E = \frac{1}{4\pi\varepsilon_0} \left(\frac{q}{r^2}\right)$, Here q is the total charge

The difference between its permittivity of the medium (ε) and the permittivity (ε0) of the free space

The reciprocal of the product of its permittivity of the medium (ε) and the permittivity (ε0) of the free space

The ratio between its permittivity of the medium (ε) and the permittivity (ε0) of the free space (correct)

The product of its permittivity of the medium (ε) and the permittivity (ε0) of the free space

$E = \frac{\sigma}{2\varepsilon_0}$ (correct)

E = 2\pi\sigma\varepsilon_0

E = 4\pi\sigma\varepsilon_0

$E = \frac{\sigma}{\varepsilon_0}$

<p>The charge density of the sheet</p> Signup and view all the answers

<p>As the negative line integral of electric field from infinity to that point along any path</p> Signup and view all the answers