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Electric Field - Practice Questions & MCQ
Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main
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Electric field is considered one of the most asked concept.
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71 Questions around this concept.
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If $E$ is the electric field intensity of an electrostatic field, then the electrostatic energy density is proportional to
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
ABC is an equilateral triangle. Charges +q are placed at each corner. The electric intensity at O will be
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Two point charges +8q and -2q are located at x = 0 and x = L respectively. The location of a point on the x axis at which the net electric field due to these two point charges is zero is
The bob of a simple pendulum has a mass of 2 g and a charge of $5.0 \mu C$. It is at rest in a uniform horizontal electric field of intensity $2000 \mathrm{~V} / \mathrm{m}$. At equilibrium, the angle that the pendulum makes with the vertical is : (take $g=10 \mathrm{~m} / \mathrm{s}^2$ ).
What is the magnitude of a point charge due to which the electric field 30 cm away has the magnitude 2 Newton/coulomb ?
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
A long cylindrical shell carries positive surface charge $\sigma$ in the upper half and negative surface charge $-\sigma$ in the lower half. The electric field lines around the cylinder will look like figure given in : (figures are schematic and not drawn to scale)
The unit of intensity of electric field is:
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Download EBookCharges are placed on the vertices of a square as shown. Let $\vec{E}$ be the electric field and V the potential at the centre. If the charges on A and B are interchanged with those on D and C respectively, then
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 q 0 . The Sl unit of E is,
$
\frac{\text { Newton }}{\text { Columb }}=\frac{\text { Volt }}{\text { meter }}=\frac{\text { Joule }}{\text { Coulomb } \times \text { Meter }}
$
The dimensional formula is $\left[M L T^{-3} A^{-1}\right]$
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{q q_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{q q_0}{r^2} \hat{r}=\frac{1}{4 \pi \epsilon_0} \frac{q}{r^2} \hat{r}
$
(As $\mathrm{q}_0$ tends to zero the electric field produced by q is not affected by $\mathrm{q}_0$.)
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.
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