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A particle that carries a charge ‘-q’ is placed at rest in uniform electric field 10 N/C. It experiences a force and moves. In a certain time ‘t’, it is observed to acquire a velocity 10hati - 10hatj m/s. The given electric field intersects a surface of area1 m^2 in the x-z plane. Find the Electric flux through the surface. |
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Answer» Solution :Force on CHARGE `BAR(F)=qbar(E)` `:.` particle moves opposite to `bar(E )` with `vec(V)` UNIT vector in the direction of `bar(V)` is `(bar(i))/(sqrt2) - (barj)/(sqrt2)` unit vector in the direction of `barE` is `(bari)/(sqrt2) - (barj)/(sqrt2)` `bar(E) = 10 [(-i)/(sqrt2) + j/(sqrt2)] ` ie., `barA = 1 xx barj` Electric FLUX `phi = barE . barA = 5sqrt(2) Nm^2//C` |
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