Electric Field and Potential: Problem 1.11, 1.12 & 1.13 with detailed Solutions

1.11) A charged particle of mass m and charge q is released from rest in a uniform electric field E. Neglecting the effect of gravity the kinetic energy attained by the particle after time 't' seconds is
    A) Eqm/t
    B) E²q²t²/2m
    C) 2E²t²/mq
    D) Eq²m/2t²

The charged particle, being acted upon by a uniform electric field, will undergo a uniformly accelerated motion.

For the particle, its initial velocity u = 0 and acceleration a = qE/m. Let after time t, v be the velocity acquired then an application of v = u + at yields v = 0 + (qE/m)t = qEt/m. In terms of kinetic energy it is (1/2)mv² = q²E²t²/2m.

Hence the kinetic energy attained by the particle as a function of time t is K(t) = q²E²t²/2m.


1.12) Six point charges are arranged at the corners of a regular hexagon as shown in the below shown image.
For JEE Mains and Advanced
If an electron is placed at the centre of the hexagon (i.e. at point O), force on it will be

    A) zero
    B) along OF
    C) along OC
    D) None of these



Try to obtain the direction of resultant electric field vector at point O. The equal point charges located at points F and C (which is a pair of opposite vertices) cancels each other's field at the point O (as O is the midpoint of the segment FC).

Now we are left with two pair of opposite vertices (A,D and B,E). The charges placed at these vertices only contribute to the resultant field at centre of the hexagon.


Electric field at point O due to the charges at vertices A and D equals (kq/a²) + (2kq/a²) = 3kq/a² directed from O to A. [Here a = OA = OB = OC = OD = OE = OF]


Similarly, the field at point O due to the charges at vertices B and E is (3kq/a²) - (kq/a²) = 2kq/a² directed from O to E.


On vectorially combining the individual fields due to the charges at each pair of opposite vertices, we get that the resultant field vector at point O lies in between the region FOA. When an electron is placed at the point O, it experiences an electrostatic force in a direction anti-parallel to the resultant electric field. This means that the net force vector on the electron lies in between the region COD.

∴ Option D is the correct choice.





1.13) A charged particle having some mass rests in equilibrium at a height h above the centre of a uniformly charged nonconducting horizontal ring of radius R. The force of gravity acts vertically downwards. The equilibrium of the particle will be stable :
    A) for all values of h
    B) only if h > R/√2
    C) only if h < R/√2
    D) only if h = R/√2
The above graph represents the nature of electric field produced by a uniformly charged circular ring along its axis. You can observe that the intensity of field is zero at the centre. On moving away from the centre, field strength rises (but with decreasing rate of rise or with decreasing slope), reaches a maximum value (this occurs at the point |x|= R/√2) and then starts decreasing. For more detailed explanations you may visit this article: Electric Field due to a uniformly charged Ring: A Comprehensive Analysis.

Furthermore, it is also observable from the graph that at points x = x₁ and x = x₂, field strength is same (though the graph is not symmetrical about |x|= R/√2). So, for the equilibrium condition of a given point mass, there may be two possibilities. It may stay at equilibrium, either at a height x < R/√2 or x > R/√2. We have to check the stability of the equilibrium in both the cases.






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