Electric potential at any point is given by V | vedclass

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Question

(D) The electric field $\vec{E}$ is related to the electric potential $V$ by the relation $\vec{E} = -
abla V = -\left( \frac{\partial V}{\partial x}

Vedclass · Accepted Answer

$7$

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(D) The electric field $\vec{E}$ is related to the electric potential $V$ by the relation $\vec{E} = -
abla V = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} ight)$.Given $V = -5x + 3y + \sqrt{15}z$.Calculating the partial derivatives:$E_x = -\frac{\partial V}{\partial x} = -(-5) = 5$$E_y = -\frac{\partial V}{\partial y} = -(3) = -3$$E_z = -\frac{\partial V}{\partial z} = -(\sqrt{15}) = -\sqrt{15}$Thus,$\vec{E} = 5\hat{i} - 3\hat{j} - \sqrt{15}\hat{k}$.The magnitude of the electric field is $|\vec{E}| = \sqrt{E_x^2 + E_y^2 + E_z^2}$.$|\vec{E}| = \sqrt{(5)^2 + (-3)^2 + (-\sqrt{15})^2} = \sqrt{25 + 9 + 15} = \sqrt{49} = 7$.

Electric potential at any point is given by $V = -5x + 3y + \sqrt{15}z$. The magnitude of the electric field is:

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