a particle of mass m and charge q is subjected to combined action of gravity and a uniform horizontal electric field of strength E. it is projected from ground with speed v in the vertical plane parallel to the field at an angle N to the horizontal . what is the maximum distance the object can travel horizontally before striking ground. the ans is v^2(qE +square root of (m^2g^2 +q^2E^2)) /mg^2

Question

Sher Mohammad · Answer

force on particle
F= qe i+-mg kintial velocity, u=VcosN i+ VsinN k
at maximum height vertical velocity is zero,
0=VsinN- mgt=0
t=V SinN/mg
the time to come back to ground = 2t= 2VSinN/mg
distance travelled , s=ut+.5 a t^2= VcosN*2VSinN/mg+ .5*qE*(2VSinN/mg)^2

sher mohammad
b.tech, iit delhi

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