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The electirc potential at a point (x, y,...

The electirc potential at a point `(x, y, z)` is given by
`V = -x^(2)y - xz^(3) + 4`
The electric field `vecE` at that point is

A

`vecE= hati (2xy -z^(3)) +hatj xy^(2) +hatk 3z^(2)x`

B

`vecE = hati (2xy +z^(3)) +hatj x^(2) +hatk 3xz^(2)`

C

`vecE =hati 2xy +hatj (x^(2)+y^(2)) +hatk (3xz -y^(2))`

D

`vecE =hati z^(3) +hatj xyz +hatk z^(2)`

Text Solution

AI Generated Solution

The correct Answer is:
To find the electric field \(\vec{E}\) at a point \((x, y, z)\) given the electric potential \(V\), we can use the relationship between electric potential and electric field. The electric field is given by the negative gradient of the electric potential: \[ \vec{E} = -\nabla V \] This means we need to calculate the partial derivatives of \(V\) with respect to \(x\), \(y\), and \(z\). ### Step 1: Calculate \(E_x\) The \(x\)-component of the electric field \(E_x\) is given by: \[ E_x = -\frac{\partial V}{\partial x} \] Given \(V = -x^2 y - x z^3 + 4\), we differentiate \(V\) with respect to \(x\): \[ \frac{\partial V}{\partial x} = -2xy - z^3 \] Thus, \[ E_x = -(-2xy - z^3) = 2xy + z^3 \] ### Step 2: Calculate \(E_y\) The \(y\)-component of the electric field \(E_y\) is given by: \[ E_y = -\frac{\partial V}{\partial y} \] We differentiate \(V\) with respect to \(y\): \[ \frac{\partial V}{\partial y} = -x^2 \] Thus, \[ E_y = -(-x^2) = x^2 \] ### Step 3: Calculate \(E_z\) The \(z\)-component of the electric field \(E_z\) is given by: \[ E_z = -\frac{\partial V}{\partial z} \] We differentiate \(V\) with respect to \(z\): \[ \frac{\partial V}{\partial z} = -3xz^2 \] Thus, \[ E_z = -(-3xz^2) = 3xz^2 \] ### Step 4: Combine the components into the electric field vector Now we can write the electric field vector \(\vec{E}\): \[ \vec{E} = E_x \hat{i} + E_y \hat{j} + E_z \hat{k} \] Substituting the values we found: \[ \vec{E} = (2xy + z^3) \hat{i} + (x^2) \hat{j} + (3xz^2) \hat{k} \] ### Final Answer Thus, the electric field \(\vec{E}\) at the point \((x, y, z)\) is: \[ \vec{E} = (2xy + z^3) \hat{i} + (x^2) \hat{j} + (3xz^2) \hat{k} \] ---
To find the electric field \(\vec{E}\) at a point \((x, y, z)\) given the electric potential \(V\), we can use the relationship between electric potential and electric field. The electric field is given by the negative gradient of the electric potential: \[ \vec{E} = -\nabla V \] This means we need to calculate the partial derivatives of \(V\) with respect to \(x\), \(y\), and \(z\). ...
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