The electric field strenth vector is given by E=`2.5x hat I + 1.5y hat`.The potential at point (2,2,1) is (considering potential at origin to be zero)

Options

A. `-2` V
B. `-4` V
C. `-6` V
D. `-8` V

Answer

To find the electric potential at the point (2, 2, 1) given the electric field strength vector \( \mathbf{E} = 2.5 \hat{i} + 1.5 \hat{j} \), we will follow these steps:

### Step 1: Understand the relationship between electric field and potential
The relationship between the electric field \( \mathbf{E} \) and the electric potential \( V \) is given by the equation:
\[ dV = -\mathbf{E} \cdot d\mathbf{r} \]
where \( d\mathbf{r} = dx \hat{i} + dy \hat{j} + dz \hat{k} \).

### Step 2: Write down the electric field and the differential displacement
Given:
\[ \mathbf{E} = 2.5 \hat{i} + 1.5 \hat{j} \]
The differential displacement vector is:
\[ d\mathbf{r} = dx \hat{i} + dy \hat{j} + dz \hat{k} \]

### Step 3: Calculate the dot product
Now, we calculate the dot product \( \mathbf{E} \cdot d\mathbf{r} \):
\[ \mathbf{E} \cdot d\mathbf{r} = (2.5 \hat{i} + 1.5 \hat{j}) \cdot (dx \hat{i} + dy \hat{j} + dz \hat{k}) = 2.5 dx + 1.5 dy \]

### Step 4: Substitute into the equation for potential
Substituting the dot product into the equation for \( dV \):
\[ dV = - (2.5 dx + 1.5 dy) \]

### Step 5: Integrate to find the potential function
To find the potential \( V \), we integrate:
\[ V = -\int (2.5 dx + 1.5 dy) \]
This can be separated into two integrals:
\[ V = -\left( \int 2.5 dx + \int 1.5 dy \right) \]
Integrating each term:
\[ V = -\left( 2.5 x + 1.5 y \right) + C \]
where \( C \) is the integration constant.

### Step 6: Determine the constant using the reference point
We know that the potential at the origin (0, 0, 0) is zero:
\[ V(0, 0, 0) = -\left( 2.5 \cdot 0 + 1.5 \cdot 0 \right) + C = 0 \implies C = 0 \]
Thus, the potential function simplifies to:
\[ V = -\left( 2.5 x + 1.5 y \right) \]

### Step 7: Calculate the potential at the point (2, 2, 1)
Now, we substitute \( x = 2 \) and \( y = 2 \):
\[ V(2, 2, 1) = -\left( 2.5 \cdot 2 + 1.5 \cdot 2 \right) \]
Calculating this:
\[ V(2, 2, 1) = -\left( 5 + 3 \right) = -8 \text{ volts} \]

### Final Answer
The potential at the point (2, 2, 1) is:
\[ \boxed{-8 \text{ volts}} \]