Can the expression E = xyî + 2yzĵ + 3xz k be said to be that of an electrostatic field? Justify your answer.

Physics
Can the expression E = xyî + 2yzĵ + 3xz k be said to be that of an electrostatic field? Justify your answer.

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Answer

No, the expression does not represent an electrostatic field.

To determine if the given expression represents an electrostatic field, we need to check if its curl is zero. An electrostatic field is a conservative field, and a fundamental property of conservative fields is that their curl is zero (∇×E⃗=0\nabla \times \vec{E} = 0).

The given electric field is: E⃗=xyi^+2yzj^+3xzk^\vec{E} = xy\hat{i} + 2yz\hat{j} + 3xz\hat{k} Here, Ex=xyE_x = xy, Ey=2yzE_y = 2yz, and Ez=3xzE_z = 3xz.

Step 1: Calculate the curl of the electric field. The curl operator is given by: ∇×E⃗=(∂Ez∂y−∂Ey∂z)i^+(∂Ex∂z−∂Ez∂x)j^+(∂Ey∂x−∂Ex∂y)k^\nabla \times \vec{E} = \left( \frac{\partial E_z}{\partial y} - \frac{\partial E_y}{\partial z} \right)\hat{i} + \left( \frac{\partial E_x}{\partial z} - \frac{\partial E_z}{\partial x} \right)\hat{j} + \left( \frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial y} \right)\hat{k}

Step 2: Compute the partial derivatives for each component. For the i^\hat{i} component: ∂Ez∂y=∂∂y(3xz)=0\frac{\partial E_z}{\partial y} = \frac{\partial}{\partial y}(3xz) = 0 ∂Ey∂z=∂∂z(2yz)=2y\frac{\partial E_y}{\partial z} = \frac{\partial}{\partial z}(2yz) = 2y So, the i^\hat{i} component is 0−2y=−2y0 - 2y = -2y.

For the j^\hat{j} component: ∂Ex∂z=∂∂z(xy)=0\frac{\partial E_x}{\partial z} = \frac{\partial}{\partial z}(xy) = 0 ∂Ez∂x=∂∂x(3xz)=3z\frac{\partial E_z}{\partial x} = \frac{\partial}{\partial x}(3xz) = 3z So, the j^\hat{j} component is 0−3z=−3z0 - 3z = -3z.

For the k^\hat{k} component: ∂Ey∂x=∂∂x(2yz)=0\frac{\partial E_y}{\partial x} = \frac{\partial}{\partial x}(2yz) = 0 ∂Ex∂y=∂∂y(xy)=x\frac{\partial E_x}{\partial y} = \frac{\partial}{\partial y}(xy) = x So, the k^\hat{k} component is 0−x=−x0 - x = -x.

Step 3: Combine the components to find the curl. ∇×E⃗=−2yi^−3zj^−xk^\nabla \times \vec{E} = -2y\hat{i} - 3z\hat{j} - x\hat{k}

Step 4: Justify the answer. Since ∇×E⃗=−2yi^−3zj^−xk^\nabla \times \vec{E} = -2y\hat{i} - 3z\hat{j} - x\hat{k} is not equal to zero, the given expression for E⃗\vec{E} does not represent an electrostatic field. An electrostatic field must have a curl of zero because it is a conservative field, meaning the line integral of the electric field around any closed loop is zero.

The expression below cannot be said to be that of an electrostatic field. Justification: An electrostatic field is a conservative field, which implies that its curl must be zero. The calculated curl of the given field is ∇×E⃗=−2yi^−3zj^−xk^\nabla \times \vec{E} = -2y\hat{i} - 3z\hat{j} - x\hat{k}, which is not zero.

The final answer is No,theexpressiondoesnotrepresentanelectrostaticfield.\boxed{No, the expression does not represent an electrostatic field.}

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To determine if the given expression represents an electrostatic field, we need to check if its curl is zero.

Can the expression E = xyî + 2yzĵ + 3xz k be said to be that of an electrostatic field? Justify your answer.
Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

ScanSolve AI Answer
To determine if the given expression represents an electrostatic field, we need to check if its curl is zero. An electrostatic field is a conservative field, and a fundamental property of conservative fields is that their curl is zero ( × E = 0). The given electric field is: E = xyi + 2yzj + 3xzk Here, E_x = xy, E_y = 2yz, and E_z = 3xz. Step 1: Calculate the curl of the electric field. The curl operator is given by: × E = ( ( E_z)/( y) - ( E_y)/( z) )i + ( ( E_x)/( z) - ( E_z)/( x) )j + ( ( E_y)/( x) - ( E_x)/( y) )k Step 2: Compute the partial derivatives for each component. For the i component: ( E_z)/( y) = ()/( y)(3xz) = 0 ( E_y)/( z) = ()/( z)(2yz) = 2y So, the i component is 0 - 2y = -2y. For the j component: ( E_x)/( z) = ()/( z)(xy) = 0 ( E_z)/( x) = ()/( x)(3xz) = 3z So, the j component is 0 - 3z = -3z. For the k component: ( E_y)/( x) = ()/( x)(2yz) = 0 ( E_x)/( y) = ()/( y)(xy) = x So, the k component is 0 - x = -x. Step 3: Combine the components to find the curl. × E = -2yi - 3zj - xk Step 4: Justify the answer. Since × E = -2yi - 3zj - xk is not equal to zero, the given expression for E does not represent an electrostatic field. An electrostatic field must have a curl of zero because it is a conservative field, meaning the line integral of the electric field around any closed loop is zero. The expression below cannot be said to be that of an electrostatic field. Justification: An electrostatic field is a conservative field, which implies that its curl must be zero. The calculated curl of the given field is × E = -2yi - 3zj - xk, which is not zero. The final answer is No, the expression does not represent an electrostatic field.