Find out the electric flux through an area `10 m^(2)` lying in XY plane due to a electric field `vec(E)=2hat(i)-10 hat(j)+5hat(k)`.

Answer

To find the electric flux through an area of \(10 \, m^2\) lying in the XY plane due to the electric field \(\vec{E} = 2\hat{i} - 10\hat{j} + 5\hat{k}\), we can follow these steps:

### Step 1: Understand the concept of electric flux
Electric flux (\(\Phi\)) through a surface is given by the formula:
\[ \Phi = \vec{E} \cdot \vec{A} \]
where \(\vec{A}\) is the area vector, and \(\vec{E}\) is the electric field vector.

### Step 2: Determine the area vector
Since the area is lying in the XY plane, the area vector \(\vec{A}\) will be perpendicular to the XY plane, which means it will be in the direction of the Z-axis. Therefore, we can write:
\[ \vec{A} = A \hat{k} = 10 \hat{k} \, m^2 \]

### Step 3: Write the electric field vector
The electric field is given as:
\[ \vec{E} = 2\hat{i} - 10\hat{j} + 5\hat{k} \]

### Step 4: Calculate the dot product
Now, we can calculate the electric flux by taking the dot product of \(\vec{E}\) and \(\vec{A}\):
\[ \Phi = \vec{E} \cdot \vec{A} = (2\hat{i} - 10\hat{j} + 5\hat{k}) \cdot (10\hat{k}) \]

### Step 5: Perform the dot product
Using the properties of the dot product:
\[ \Phi = 2\hat{i} \cdot 10\hat{k} - 10\hat{j} \cdot 10\hat{k} + 5\hat{k} \cdot 10\hat{k} \]
Since \(\hat{i} \cdot \hat{k} = 0\) and \(\hat{j} \cdot \hat{k} = 0\), we only need to consider the last term:
\[ \Phi = 0 - 0 + 50 = 50 \, \text{N m}^2/\text{C} \]

### Final Answer
Thus, the electric flux through the area is:
\[ \Phi = 50 \, \text{N m}^2/\text{C} \]

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