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We know that electric field (E ) at any point in space can be calculated using the relation `vecE = - (deltaV)/(deltax)hati - (deltaV)/(deltay)hatj - (deltaV)/(deltaz)hatk` if we know the variation of potential (V) at that point. Now let the electric potential in volt along the x-axis vary as `V = 2x^2`, where x is in meter. Its variation is as shown in figure A charge particle of mass 10 mg and charge `2.5 muC` is released from rest at `x = 2 m`. Find its velocity when it crosses origin.A. `0.5ms^(-1)`B. `1 ms^(-1)`C. `2ms^(-1)`D. `4 ms^(-1)` |
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Answer» Correct Answer - C `E=-(dV)/(dx)=-4x` `F=qE=2.5xx10^(-6)(-4x)=-10^(-5)x` `W=underset(2)overset(0)int Fdx=-10^(-5)underset(2)overset(0)intxdx` `(1)/(2)mv^(2)=10^(-5)[(x^(2))/(2)]_(0)^(2)` or `V=2ms^(-1)` |
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