If a proton is moved against the Coulomb force of an electric field,then

  • A
    work is done by the electric field
  • B
    energy is used from some outside source
  • C
    the strength of the field is decreased
  • D
    the strength of the field is increased

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Similar Questions

$A$ uniform electric field,$\vec{E} = -400 \sqrt{3} \hat{y} \text{ NC}^{-1}$ is applied in a region. $A$ charged particle of mass $m$ carrying positive charge $q$ is projected in this region with an initial speed of $u = 2 \sqrt{10} \times 10^6 \text{ ms}^{-1}$. This particle is aimed to hit a target $T$,which is $5 \text{ m}$ away from its entry point into the field as shown schematically in the figure. Take $\frac{q}{m} = 10^{10} \text{ Ckg}^{-1}$. Then-
$(A)$ the particle will hit $T$ if projected at an angle $45^{\circ}$ from the horizontal
$(B)$ the particle will hit $T$ if projected either at an angle $30^{\circ}$ or $60^{\circ}$ from the horizontal
$(C)$ time taken by the particle to hit $T$ could be $\sqrt{\frac{5}{6}} \mu\text{s}$ as well as $\sqrt{\frac{5}{2}} \mu\text{s}$
$(D)$ time taken by the particle to hit $T$ is $\sqrt{\frac{5}{3}} \mu\text{s}$

The figures below depict two situations in which two infinitely long static line charges of constant positive line charge density $\lambda$ are kept parallel to each other. In their resulting electric field, point charges $+q$ and $-q$ are kept in equilibrium between them. The point charges are confined to move in the $x$ direction only. If they are given a small displacement about their equilibrium positions, then the correct statement(s) is(are):

In a uniform electric field of $E = 1 \times 10^{4} \text{ N C}^{-1}$,an electron is accelerated from rest. The velocity of the electron when it has travelled a distance of $2 \times 10^{-2} \text{ m}$ is nearly (given $\frac{e}{m} \text{ of electron } \approx 1.8 \times 10^{11} \text{ C kg}^{-1}$):

An oil drop having a mass $4.8 \times 10^{-13} \,kg$ and charge $2.4 \times 10^{-18} \,C$ stands still between two charged horizontal plates separated by a distance of $1 \,cm$. If now the polarity of the plates is changed,the instantaneous acceleration of the drop is: $(g = 10 \,m/s^2)$ (in $\,m/s^2$)

$A$ small sphere of mass $m = 0.5 \, kg$ carrying a positive charge $q = 110 \, \mu C$ is connected to a light,flexible,and inextensible string of length $r = 60 \, cm$ and whirled in a vertical circle. If a vertically upwards electric field of strength $E = 10^5 \, N/C$ exists in the space,what is the minimum velocity of the sphere required at the highest point so that it may just complete the circle? $(g = 10 \, m/s^2)$

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