The electric field due to a charge at a dista | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The electric field $E$ due to a point charge $Q$ at a distance $r$ is given by the formula: $E = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Q}{r^2}$.

Vedclass · Accepted Answer

$0.5$

Vedclass · Answer

(D) The electric field $E$ due to a point charge $Q$ at a distance $r$ is given by the formula: $E = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Q}{r^2}$.Given values are $E = 500\, N/C$,$r = 3\, m$,and $k = \frac{1}{4\pi\varepsilon_0} = 9 	imes 10^9\, N\cdot m^2/C^2$.Substituting these values into the formula:$500 = (9 	imes 10^9) \cdot \frac{Q}{3^2}$$500 = (9 	imes 10^9) \cdot \frac{Q}{9}$$500 = 10^9 \cdot Q$$Q = \frac{500}{10^9} = 5 	imes 10^{-7}\, C$.To convert this into microcoulombs $(\mu C)$,we multiply by $10^6$:$Q = 5 	imes 10^{-7} 	imes 10^6\, \mu C = 0.5\, \mu C$.

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

Similar Questions

Explain electric field and its source as well as magnetic field and its source.

The given diagram shows two semi-infinite lines of charge having equal (in magnitude) linear charge density but with opposite signs. The electric field at any point on the $x$-axis for $(x > 0)$ is along the unit vector:

What is called an electric field?

Is electric field a scalar or a vector quantity? Why?

The distance between a proton and an electron in a hydrogen atom is ${10^{ - 10}} \ m$. Both have a charge of magnitude $1.6 \times {10^{ - 19}} \ C$. The magnitude of the electric field intensity produced at the position of the electron due to the proton is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module