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# Electric Field - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• Electric field is considered one of the most asked concept.

• 62 Questions around this concept.

## Solve by difficulty

If a charge  q is placed at the centre of the line joining two equal charges Q  such that the system is in equilibrium then the value of q is

What is the magnitude of a point charge due to which the electric field 30 cm away has the magnitude 2 Newton/coulomb $[1/4\pi \varepsilon _{0}=9\times 10^{9}Nm^{2}]$ ?

Two point charges Q and – 3Q are placed at some distance apart. If the electric field at the location of Q is E, then at the locality of – 3Q, it is

## Concepts Covered - 1

Electric field

Electric field and electric lines of force:

The space around a charge in which another charged particle experiences a force is said to have an electric field in it.

Electric Field Intensity:

The electric field intensity at any point is defined as the force experienced by a unit positive charge placed at that point.

Electric field intensity,

$E=\frac{F}{q_0}$

where F is the force experienced by q0. The SI unit of E is,

$\frac{Newton}{Columb}=\frac{Volt}{meter}=\frac{Joule}{Coulomb\times Meter}$

The dimensional formula is $[MLT^{-3}A^{-1}]$

The electric field is a vector quantity and due to positive charge is away from the charge and for the negative charge, it is towards the charge.

Electric field due to a point charge:

Consider a point charge placed at the origin O. Let a test charge q0 is placed at P which is at a distance r from O. Force F on test charge q0 is

$F=\frac{1}{4\pi \epsilon_0}\frac{qq_0}{r^2}\hat r$

The electric field at point P due to q is

$E=\lim_{q_0\rightarrow 0}\frac{F}{q_0}=\lim_{q_0\rightarrow 0}\frac{1}{q_0}\frac{1}{4\pi \epsilon_0}\frac{qq_0}{r^2}\hat r=\frac{1}{4\pi \epsilon_0}\frac{q}{r^2}\hat r$

(As qo tends to zero the electric field produced by q is not affected by qo.)

The magnitude of the electric field

$E=\frac{1}{4\pi \epsilon_0}\frac{q}{r^2}$

Electric field due to a system of charge:

Electric field obeys the superposition principle. That is the electric field due to a system of charge at a point is equal to the vector sum of all the electric fields.

## Study it with Videos

Electric field

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