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: b
Explanation: V = -∫ E.dl = -∫ (40xy dx + 20×2 dy + 2 dz), from (2,1,3) to (1,-1,0), we get
Vpq on integrating from Q to P. Vpq = 106 volts
To find the potential difference between two points given an electric field vector ( mathbf{E} = 40xymathbf{i} + 20x^2 mathbf{j} + 2mathbf{k} ), we recognize this requires integrating the electric field along a path from point A ((1, -1, 0)) to point B ((2, 1, 3)). The potential difference ((V)) between two points in an electric field is found using the negative integral of the electric field along the path from point A to point B. Mathematically, this is given as:
[ V = – int_{A}^{B} mathbf{E} cdot dmathbf{r} ]
Given ( mathbf{E} = 40xymathbf{i} + 20x^2mathbf{j} + 2mathbf{k} ), let’s decompose this problem into a manageable form.
The differential length vector ( dmathbf{r} ) in Cartesian coordinates is given by
[ dmathbf{r} = dxmathbf{i} + dymathbf{j} + dzmathbf{k} ]
So, the dot product ( mathbf{E} cdot dmathbf{r} ) becomes
[ mathbf{E} cdot dmath