Electric Field

A proton and an $\alpha$ particle having equal kinetic energy are projected in a uniform transverse electric field as shown in figure

A circular metal plate of radius $1.5 \mathrm{cm}$ is charged with $10 \mathrm{\mu C}$ situated in air. If the same circular plate carrying same charge were kept in acetone, what would be magnitude of mechanical force? ($k_{\mathrm{acetone}}$ $=27$)

A proton is released from rest, $10\mathrm{cm}$ from a charged sheet carrying charged density of $-2.21\times {10}^{-9}\mathrm{C} {\mathrm{m}}^{-2}$. It will strike the sheet after the time (approximately)

An electron moving with the speed $5\times {10}^6$ per sec is shooted parallel to the electric field of intensity $1\times {10}^3 \mathrm{N} {\mathrm{C}}^{-1}$. Field is responsible for the retardation of motion of electron. Now evaluate the distance travelled by the electron before coming to rest for an instant (mass of $e=9\times {10}^{-31}\mathrm{kg}$, charge $=1.6\times {10}^{-19}\mathrm{C}$)

The Electric field is given by $\overrightarrow{E}=\frac{\overrightarrow{F}}{q_0}$, here the test charge '$q_0$' should be
a) Infinitesimally small and positive

b) Infinitesimally small and negative

Which one value of electric field is possible due to uniformly charged ring?

Electric field at point $P\left(3,4,5\right) \mathrm{m}$ due to charge $4 \mathrm{nC}$ which is placed at $Q\left(2,3,4\right) \mathrm{m}$ is 

A cube of side $b$ has a charge $q$ at seven of its vertices. The electric field due to this charge distribution at the centre of this cube will be:

In the following figure, the electric field on $y$-axis will be maximum at $y=$

Electric field strength due to a proton at a distance of $3 \mathrm{m}$ is

Electric field strength due to a proton at a distance of $3 \mathrm{m}$ is

A particle is moving in an electric field ${10}^3 \mathrm{N} {\mathrm{C}}^{-1}$ along $x$-axis, the charge $=1 \mathrm{\mu C}$ is present on particle and mass $m=1 \mathrm{g}$. A magnetic field of $10 \mathrm{T}$ is also present. The particle starts moving from origin at $t=0$ along the same axis with the velocity of $\overrightarrow{v}=20\hat{\mathrm{j}} \mathrm{m} {\mathrm{s}}^{-1}$ as shown, the speed of the particle at the time of $20\sqrt{3} \mathrm{s}$ will be:

Assertion: Due to two point charges electric field and electric potential cannot be zero at some point simultaneously.

Reason: Field is a vector quantity and potential is a scalar quantity.

Two point charges $q$ and $2q$ are placed some distance apart. If the electric field at the location of $q$ be $E$, then that at the location of $2q$ will be

Two point charges of $20 \mu \mathrm{C}$ and $80 \mu \mathrm{C}$ are $10 \mathrm{cm}$ apart where will the electric field strength be zero on the line joining the charges from $20 \mu \mathrm{C}$ charge

A charged particle of mass $5\times {10}^{-5} \mathrm{kg}$ is held stationary in space by placing it in an electric field of strength ${10}^7 \mathrm{N} {\mathrm{C}}^{-1}$ directed vertically downwards. The charge on the particle is

A small metal ball is suspended in an uniform electric field with the help of an insulated thread. If a high energy $\mathrm{X}$-ray beam falls on the ball, then the ball

An electrostatic field line leaves at an angle $\alpha$ from point charge $q_1$ and connects with point charge $-q_2$ at an angle $\beta$($q_1$and $q_2$ are positive) see figure below. If $q_2=\frac{3}{2}{q}_1$ and $\alpha =30°,$ then

Two masses $M_1$ and $M_2$ carry positive charges $Q_1$ and $Q_2,$ respectively. They are dropped to the floor in a laboratory set up from the same height, where there is a constant electric field vertically upwards. $M_1$ hits the floor before $M_2$. Then,