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Electric potential in a region of space is given by V=3x^(2)V. What will be electric field intensity at a point P (2 m, 0). |
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Answer» Solution :As we can see here that the electric potential is DEPENDENT only on ONE variable that is x-coordinate. We can calculate electric field intensity at the given POINT simply differentiating the given function with respect to x. Electric field intensity is given by: `E= -(dV)/(dx)= -(d(3x^(2)))/(dx)=-6x rArr -6(2) N//C=-12N//C` The direction of the electric field intensity will be along the negative direction of X-axis. In this question, if the y-coordinate of the given point were not 0, even then the electric field intensity would be same. We can conclude here that the planes parallel to x-z planes can be considered as equipotential surfaces. |
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