Assume that an electric field `vec(E) = 30 x^(2) hat(i)` exists in space. Then the potential differences `V_(A) - V_(0)` where `V_(0)` is the potential at the origin and `V_(A)`, the potential at `x = 2m` is

Options

A. 120 V
B. `-120 V`
C. `-80 V`
D. `80 V`

Answer

To find the potential difference \( V_A - V_0 \) where \( V_0 \) is the potential at the origin and \( V_A \) is the potential at \( x = 2 \, \text{m} \), we can follow these steps:

### Step 1: Understand the relationship between electric field and potential
The potential difference \( dV \) between two points in an electric field is given by:
\[ dV = -\vec{E} \cdot d\vec{x} \]
where \( \vec{E} \) is the electric field and \( d\vec{x} \) is the displacement vector.

### Step 2: Set up the integral for potential difference
To find the potential difference \( V_A - V_0 \), we need to integrate the expression from the origin (0) to the point at \( x = 2 \, \text{m} \):
\[ V_A - V_0 = -\int_{0}^{2} \vec{E} \cdot d\vec{x} \]

### Step 3: Substitute the electric field
Given that the electric field is \( \vec{E} = 30x^2 \hat{i} \), the displacement vector in the \( x \) direction can be expressed as \( d\vec{x} = dx \hat{i} \). Therefore, the dot product becomes:
\[ \vec{E} \cdot d\vec{x} = (30x^2 \hat{i}) \cdot (dx \hat{i}) = 30x^2 \, dx \]

### Step 4: Set up the integral
Now we can rewrite the integral:
\[ V_A - V_0 = -\int_{0}^{2} 30x^2 \, dx \]

### Step 5: Calculate the integral
Now we compute the integral:
\[ \int 30x^2 \, dx = 30 \cdot \frac{x^3}{3} = 10x^3 \]
Evaluating this from 0 to 2:
\[ V_A - V_0 = -\left[ 10x^3 \right]_{0}^{2} = -\left[ 10(2^3) - 10(0^3) \right] = -\left[ 10 \cdot 8 - 0 \right] = -80 \]

### Step 6: Final result
Thus, the potential difference \( V_A - V_0 \) is:
\[ V_A - V_0 = -80 \, \text{V} \]

### Summary
The potential difference \( V_A - V_0 \) is \( -80 \, \text{V} \).

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