Electric field due to a charge at a distance of 3, m from it is 500, N/coulomb . magnitude of charge

English

Gujarati

Hindi

1. Electric Charges and Fields

easy

The electric field due to a charge at a distance of $3\, m$ from it is $500\, N/coulomb$. The magnitude of the charge is.......$\mu C$ $\left[ {\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {{10}^9}\,\frac{{N - {m^2}}}{{coulom{b^2}}}} \right]$

A

$2.5$

B

$2.0$

C

$1.0$

D

$0.5$

Solution

(d) $E = 9 \times {10^9} \times \frac{Q}{{{r^2}}}\, \Rightarrow 500 = 9 \times {10^9} \times \frac{Q}{{{{(3)}^2}}}$ $==>$ $Q = 0.5 \,\mu C$

Standard 12

Physics

Share

Similar Questions

An oil drop carries six electronic charges, has a mass of $1.6 \times 10^{-12} g$ and falls with a terminal velocity in air. The magnitude of vertical electrical electric field required to make the drop move upward with the same speed as was formely moving is ……..$kN/C$

medium

Explain electric field and also electric field by point charge.

medium

The distance between the two charges $25\,\mu C$ and $36\,\mu C$ is $11\,cm$ At what point on the line joining the two, the intensity will be zero

medium

$(a)$ Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where $E =0$ ) of the configuration. Show that the equilibrium of the test charge is necessarily unstable.

$(b)$ Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.

medium

Four charges $q, 2q, -4q$ and $2q$ are placed in order at the four corners of a square of side $b$. The net field at the centre of the square is

hard