Ed
há 23 horas
Vamos calcular o campo elétrico E no ponto P(1,1,2), sabendo que o potencial é: \[ V = 2x^2 y + 3 y z \] O campo elétrico é dado por: \[ \vec{E} = - \nabla V = - \left( \frac{\partial V}{\partial x} \hat{a}_x + \frac{\partial V}{\partial y} \hat{a}_y + \frac{\partial V}{\partial z} \hat{a}_z \right) \] 1. Calcular as derivadas parciais: \[ \frac{\partial V}{\partial x} = \frac{\partial}{\partial x} (2x^2 y + 3 y z) = 4 x y + 0 = 4 x y \] \[ \frac{\partial V}{\partial y} = \frac{\partial}{\partial y} (2x^2 y + 3 y z) = 2 x^2 + 3 z \] \[ \frac{\partial V}{\partial z} = \frac{\partial}{\partial z} (2x^2 y + 3 y z) = 0 + 3 y = 3 y \] 2. Avaliar no ponto P(1,1,2): \[ \frac{\partial V}{\partial x} = 4 \times 1 \times 1 = 4 \] \[ \frac{\partial V}{\partial y} = 2 \times 1^2 + 3 \times 2 = 2 + 6 = 8 \] \[ \frac{\partial V}{\partial z} = 3 \times 1 = 3 \] 3. Portanto, \[ \vec{E} = - (4 \hat{a}_x + 8 \hat{a}_y + 3 \hat{a}_z) = -4 \hat{a}_x - 8 \hat{a}_y - 3 \hat{a}_z \quad \text{V/m} \] Resposta correta: alternativa D) E = -4 ax - 8 ay - 3 az V/m.
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