Question

Two small particles A and B having masses `m = 0.5 kg` each and charges `q_1 = (-155//18 muC)` and `q_2 = (+100 muC)`, respectively, are connected at the ends of a nonconducting, flexible, and inextensible string of length `r = 0.5 m`.
Particle A is fixed and B is whirled along a verticle circle with center at A. if a vertically ipward electric field of Strength `E = 1.1 xx 10^5 NC^-1` exists in the space, calculate the minimum velocity of particle B required at the highest point so that it may just complete the circle `(g = 10 ms^-2)`.
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Answer 1

Let us analyse the forces acting on charge B.
(i) Force of attraction between charges An and B
`F_1 = (1)/(4 pi epsilon_0)(q_1 q_2)/(r^2)`
=`9 xx 10^9 xx (((155)/(18) xx 10^-6)(100 xx 10^-6))/((0.5)^2)= 31 N`
(acts in radial direction, always directed toward the center)
(ii) Force due to electric field
`F_2 = q_2 E`
=`(100 xx 10^-6) xx (1.1 xx 10^5)`
=`11 N` (directed vertically upward)
(iii) Weight `W = mg = 0.5 xx 10 = 5 N` (vertically downward)
(iv) Tension T in the string, acting downward.
Since `F_1` is always directed toward the center, the critical position depends on `F_2` and W. their resultant is 6 N (vertically upward). At the critical position, this resultant must be directed toward the center. hence, the tension in the thread is minimum when particle B us at the lowest position. Considering free body diagram at this position, we have
`(m v_0^2)/r = F_1 + F_2 + T - W`
.
But for `T = 0`, we get `v_0 = sqrt (37) ms^-1`. When the particle moves from the lowest to the hight position, work is done on it by force `F_2` , however, gravitational potential energy increases and no work is done by `F_1` . Let minimum velocity required at the highest point be v. Using work energy theorem between the lowest and highest position on circle,
`W_("total") = Delta K`
or `W_("electric") + W_("gravity") = Delta K or qE (2 r) - m g (2 r)= (1)/(2)m v^2 - (1)/(2) m v_0^2`
which gives `v = sqrt(16) m s^-1`.