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The electric potential V at any point x, y, z (all in meters) in space is given by `V=4x^2` volts. The electric field at the point (1m, 0, 2m) is……………..`V//m`. |
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Answer» As electric field `E` is related to potential `V` through the relation `E=-(dV)/(dr)` `E_(x)=-(dV)/(dx)=-(d)/(dx)(4x^(2))=-8x` `E_(y)=(dV)/(dt)=-(d)/(dy)(4x^(2))=0` and `E_(z)=-(dV)/(dt)=-(d)/(dz)(4x^(2))=0` So `vec(E)=hat(i)E_(x)+hat(j)E_(y)+hat(k)E_(z)=-8xhat(i)` ltbr. i.e, it has magnitude `8V//m` and is directed along negative `x-axis` |
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